Smallest value of $a^2 + b^2 + c^2+ d^2$, given values for $(a+b)(c+d)$, $(a+c)(b+d)$, and $(a+d)(b+c)$












1















If $a$, $b$, $c$, $d$ belong to $mathbb{R}$, and
$$(a+b)(c+d)=143 qquad (a+c)(b+d)=150 qquad (a+d)(b+c)=169$$
Find the smallest possible value of
$$a^2 + b^2 + c^2+ d^2$$




I thought of adding $7$ to the first equation and make it equal the second, solving, and finding a new equation.



Do it $3$ times, keep substituting, but this is a very very very long approach if it's even an approach.






calculus linear-algebra algebra-precalculus







share|cite|improve this question





























    1















    If $a$, $b$, $c$, $d$ belong to $mathbb{R}$, and
    $$(a+b)(c+d)=143 qquad (a+c)(b+d)=150 qquad (a+d)(b+c)=169$$
    Find the smallest possible value of
    $$a^2 + b^2 + c^2+ d^2$$




    I thought of adding $7$ to the first equation and make it equal the second, solving, and finding a new equation.



    Do it $3$ times, keep substituting, but this is a very very very long approach if it's even an approach.






    calculus linear-algebra algebra-precalculus







    share|cite|improve this question



























      1












      1








      1


      1






      If $a$, $b$, $c$, $d$ belong to $mathbb{R}$, and
      $$(a+b)(c+d)=143 qquad (a+c)(b+d)=150 qquad (a+d)(b+c)=169$$
      Find the smallest possible value of
      $$a^2 + b^2 + c^2+ d^2$$




      I thought of adding $7$ to the first equation and make it equal the second, solving, and finding a new equation.



      Do it $3$ times, keep substituting, but this is a very very very long approach if it's even an approach.






      calculus linear-algebra algebra-precalculus







      share|cite|improve this question
















      If $a$, $b$, $c$, $d$ belong to $mathbb{R}$, and
      $$(a+b)(c+d)=143 qquad (a+c)(b+d)=150 qquad (a+d)(b+c)=169$$
      Find the smallest possible value of
      $$a^2 + b^2 + c^2+ d^2$$




      I thought of adding $7$ to the first equation and make it equal the second, solving, and finding a new equation.



      Do it $3$ times, keep substituting, but this is a very very very long approach if it's even an approach.






      calculus linear-algebra algebra-precalculus




      calculus linear-algebra algebra-precalculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 27 '18 at 4:46









      Blue

      47.7k870151




      47.7k870151










      asked Nov 26 '18 at 19:20









      Ahmed I. Elsayed

      1134




      1134






















          2 Answers
          2






          active

          oldest

          votes


















          3














          Let $lambda = a + b + c + d$. When you sum over the 3 equations.



          $$(a+b)(c+d)=143,quad (a+c)(b+d)=150,quad (a+d)(b+c)=169tag{*1}$$



          LHS sums to
          $$2(ab+ac+ad+bc+bd+cd) = (a+b+c+d)^2 - (a^2+b^2+c^2+d^2)$$
          while RHS sums to $462$. This leads to
          $$a^2 + b^2 + c^2 + d^2 = lambda^2 - 462$$
          To minimize $a^2+b^2+c^2+d^2$, one just need to minimize $lambda^2$. If you look at LHS
          of the 3 equations, all of them is a product of $2$ factors which sum to $lambda$.
          In general, if we have $p + q = lambda$, then
          $$4pq = (p+q)^2 - (p-q)^2 le (p+q)^2 = lambda^2$$
          The set of 3 equations tell us



          $$begin{align}lambda^2 &ge 4max{ (a+b)(c+d), (a+c)(b+d), (a+d)(b+c) }\
          &= 4max{ 143, 150, 169 }\&= 676end{align}$$



          As a result,
          $$a^2 + b^2 + c^2 + d^2 = lambda^2 - 462 ge 214$$



          To see $214$ is the actual minimum, we need to find $(a,b,c,d)$ which satisfies $(*1)$ and $lambda^2 = 676 = 26^2$.



          Flipping all the signs of $a,b,c,d$ if necessary, we can assume $lambda = 26$.



          The third equation $(a+d)(b+c) = 169 = 13^2 = frac14 (26)^2$ tell us
          $a + d = b + c$.

          Introduce $u,v$ such that



          $$(a,b,c,d) = left( frac{13+u}{2}, frac{13+v}{2}, frac{13-v}{2}, frac{13-u}{2}right)$$



          and substitute into the first and second equation and simplify, we obtain



          $$left(frac{u+v}{2}right)^2 = 26quadtext{ and }quad left(frac{u-v}{2}right)^2 = 19$$



          Using this, we find following 4-tuple



          $$(a,b,c,d) = {small left(
          frac{13+sqrt{26}+sqrt{19}}{2},
          frac{13+sqrt{26}-sqrt{19}}{2},
          frac{13-sqrt{26}+sqrt{19}}{2},
          frac{13-sqrt{26}-sqrt{19}}{2}
          right)}$$
          is a solution of the 3 equations in $(*1)$ with $a + b + c + d = 26$.
          One can verify $a^2 + b^2 + c^2 + d^2 = 214$ for this particular solution. As a result, the lower bound $214$ is achievable and $214$ is the smallest possible value we seek.






          share|cite|improve this answer































            1














            Using A.M.-G.M. we get: $$169+143+150=(a+b)(c+d)+(a+c)(b+d)+(a+d)(b+c)=$$



            $$2ac+2ad+2bc+2bd+2ab+2cd leq 3a^2+3b^2+3c^2+3d^2$$



            So $$a^2+b^2+c^2+d^2geq 154$$



            But, unfortenuly this is not the minumum, since variables can not be all the same.






            share|cite|improve this answer























            • Cant understand why there's an inquality there
              – Ahmed I. Elsayed
              Nov 26 '18 at 21:19










            • $2xyleq x^2+y^2$........
              – greedoid
              Nov 26 '18 at 21:20










            • 2xy <= x^2 + y^^2 ? not xy <= y^2 + x^2 ?
              – Ahmed I. Elsayed
              Nov 26 '18 at 21:24










            • First.............
              – greedoid
              Nov 26 '18 at 21:25










            • I don't know how your answer addresses the question. You just show a lower bound, not the minimum.
              – Le Anh Dung
              Nov 27 '18 at 4:39













            Your Answer





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            2 Answers
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            active

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            2 Answers
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            active

            oldest

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            active

            oldest

            votes






            active

            oldest

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            3














            Let $lambda = a + b + c + d$. When you sum over the 3 equations.



            $$(a+b)(c+d)=143,quad (a+c)(b+d)=150,quad (a+d)(b+c)=169tag{*1}$$



            LHS sums to
            $$2(ab+ac+ad+bc+bd+cd) = (a+b+c+d)^2 - (a^2+b^2+c^2+d^2)$$
            while RHS sums to $462$. This leads to
            $$a^2 + b^2 + c^2 + d^2 = lambda^2 - 462$$
            To minimize $a^2+b^2+c^2+d^2$, one just need to minimize $lambda^2$. If you look at LHS
            of the 3 equations, all of them is a product of $2$ factors which sum to $lambda$.
            In general, if we have $p + q = lambda$, then
            $$4pq = (p+q)^2 - (p-q)^2 le (p+q)^2 = lambda^2$$
            The set of 3 equations tell us



            $$begin{align}lambda^2 &ge 4max{ (a+b)(c+d), (a+c)(b+d), (a+d)(b+c) }\
            &= 4max{ 143, 150, 169 }\&= 676end{align}$$



            As a result,
            $$a^2 + b^2 + c^2 + d^2 = lambda^2 - 462 ge 214$$



            To see $214$ is the actual minimum, we need to find $(a,b,c,d)$ which satisfies $(*1)$ and $lambda^2 = 676 = 26^2$.



            Flipping all the signs of $a,b,c,d$ if necessary, we can assume $lambda = 26$.



            The third equation $(a+d)(b+c) = 169 = 13^2 = frac14 (26)^2$ tell us
            $a + d = b + c$.

            Introduce $u,v$ such that



            $$(a,b,c,d) = left( frac{13+u}{2}, frac{13+v}{2}, frac{13-v}{2}, frac{13-u}{2}right)$$



            and substitute into the first and second equation and simplify, we obtain



            $$left(frac{u+v}{2}right)^2 = 26quadtext{ and }quad left(frac{u-v}{2}right)^2 = 19$$



            Using this, we find following 4-tuple



            $$(a,b,c,d) = {small left(
            frac{13+sqrt{26}+sqrt{19}}{2},
            frac{13+sqrt{26}-sqrt{19}}{2},
            frac{13-sqrt{26}+sqrt{19}}{2},
            frac{13-sqrt{26}-sqrt{19}}{2}
            right)}$$
            is a solution of the 3 equations in $(*1)$ with $a + b + c + d = 26$.
            One can verify $a^2 + b^2 + c^2 + d^2 = 214$ for this particular solution. As a result, the lower bound $214$ is achievable and $214$ is the smallest possible value we seek.






            share|cite|improve this answer




























              3














              Let $lambda = a + b + c + d$. When you sum over the 3 equations.



              $$(a+b)(c+d)=143,quad (a+c)(b+d)=150,quad (a+d)(b+c)=169tag{*1}$$



              LHS sums to
              $$2(ab+ac+ad+bc+bd+cd) = (a+b+c+d)^2 - (a^2+b^2+c^2+d^2)$$
              while RHS sums to $462$. This leads to
              $$a^2 + b^2 + c^2 + d^2 = lambda^2 - 462$$
              To minimize $a^2+b^2+c^2+d^2$, one just need to minimize $lambda^2$. If you look at LHS
              of the 3 equations, all of them is a product of $2$ factors which sum to $lambda$.
              In general, if we have $p + q = lambda$, then
              $$4pq = (p+q)^2 - (p-q)^2 le (p+q)^2 = lambda^2$$
              The set of 3 equations tell us



              $$begin{align}lambda^2 &ge 4max{ (a+b)(c+d), (a+c)(b+d), (a+d)(b+c) }\
              &= 4max{ 143, 150, 169 }\&= 676end{align}$$



              As a result,
              $$a^2 + b^2 + c^2 + d^2 = lambda^2 - 462 ge 214$$



              To see $214$ is the actual minimum, we need to find $(a,b,c,d)$ which satisfies $(*1)$ and $lambda^2 = 676 = 26^2$.



              Flipping all the signs of $a,b,c,d$ if necessary, we can assume $lambda = 26$.



              The third equation $(a+d)(b+c) = 169 = 13^2 = frac14 (26)^2$ tell us
              $a + d = b + c$.

              Introduce $u,v$ such that



              $$(a,b,c,d) = left( frac{13+u}{2}, frac{13+v}{2}, frac{13-v}{2}, frac{13-u}{2}right)$$



              and substitute into the first and second equation and simplify, we obtain



              $$left(frac{u+v}{2}right)^2 = 26quadtext{ and }quad left(frac{u-v}{2}right)^2 = 19$$



              Using this, we find following 4-tuple



              $$(a,b,c,d) = {small left(
              frac{13+sqrt{26}+sqrt{19}}{2},
              frac{13+sqrt{26}-sqrt{19}}{2},
              frac{13-sqrt{26}+sqrt{19}}{2},
              frac{13-sqrt{26}-sqrt{19}}{2}
              right)}$$
              is a solution of the 3 equations in $(*1)$ with $a + b + c + d = 26$.
              One can verify $a^2 + b^2 + c^2 + d^2 = 214$ for this particular solution. As a result, the lower bound $214$ is achievable and $214$ is the smallest possible value we seek.






              share|cite|improve this answer


























                3












                3








                3






                Let $lambda = a + b + c + d$. When you sum over the 3 equations.



                $$(a+b)(c+d)=143,quad (a+c)(b+d)=150,quad (a+d)(b+c)=169tag{*1}$$



                LHS sums to
                $$2(ab+ac+ad+bc+bd+cd) = (a+b+c+d)^2 - (a^2+b^2+c^2+d^2)$$
                while RHS sums to $462$. This leads to
                $$a^2 + b^2 + c^2 + d^2 = lambda^2 - 462$$
                To minimize $a^2+b^2+c^2+d^2$, one just need to minimize $lambda^2$. If you look at LHS
                of the 3 equations, all of them is a product of $2$ factors which sum to $lambda$.
                In general, if we have $p + q = lambda$, then
                $$4pq = (p+q)^2 - (p-q)^2 le (p+q)^2 = lambda^2$$
                The set of 3 equations tell us



                $$begin{align}lambda^2 &ge 4max{ (a+b)(c+d), (a+c)(b+d), (a+d)(b+c) }\
                &= 4max{ 143, 150, 169 }\&= 676end{align}$$



                As a result,
                $$a^2 + b^2 + c^2 + d^2 = lambda^2 - 462 ge 214$$



                To see $214$ is the actual minimum, we need to find $(a,b,c,d)$ which satisfies $(*1)$ and $lambda^2 = 676 = 26^2$.



                Flipping all the signs of $a,b,c,d$ if necessary, we can assume $lambda = 26$.



                The third equation $(a+d)(b+c) = 169 = 13^2 = frac14 (26)^2$ tell us
                $a + d = b + c$.

                Introduce $u,v$ such that



                $$(a,b,c,d) = left( frac{13+u}{2}, frac{13+v}{2}, frac{13-v}{2}, frac{13-u}{2}right)$$



                and substitute into the first and second equation and simplify, we obtain



                $$left(frac{u+v}{2}right)^2 = 26quadtext{ and }quad left(frac{u-v}{2}right)^2 = 19$$



                Using this, we find following 4-tuple



                $$(a,b,c,d) = {small left(
                frac{13+sqrt{26}+sqrt{19}}{2},
                frac{13+sqrt{26}-sqrt{19}}{2},
                frac{13-sqrt{26}+sqrt{19}}{2},
                frac{13-sqrt{26}-sqrt{19}}{2}
                right)}$$
                is a solution of the 3 equations in $(*1)$ with $a + b + c + d = 26$.
                One can verify $a^2 + b^2 + c^2 + d^2 = 214$ for this particular solution. As a result, the lower bound $214$ is achievable and $214$ is the smallest possible value we seek.






                share|cite|improve this answer














                Let $lambda = a + b + c + d$. When you sum over the 3 equations.



                $$(a+b)(c+d)=143,quad (a+c)(b+d)=150,quad (a+d)(b+c)=169tag{*1}$$



                LHS sums to
                $$2(ab+ac+ad+bc+bd+cd) = (a+b+c+d)^2 - (a^2+b^2+c^2+d^2)$$
                while RHS sums to $462$. This leads to
                $$a^2 + b^2 + c^2 + d^2 = lambda^2 - 462$$
                To minimize $a^2+b^2+c^2+d^2$, one just need to minimize $lambda^2$. If you look at LHS
                of the 3 equations, all of them is a product of $2$ factors which sum to $lambda$.
                In general, if we have $p + q = lambda$, then
                $$4pq = (p+q)^2 - (p-q)^2 le (p+q)^2 = lambda^2$$
                The set of 3 equations tell us



                $$begin{align}lambda^2 &ge 4max{ (a+b)(c+d), (a+c)(b+d), (a+d)(b+c) }\
                &= 4max{ 143, 150, 169 }\&= 676end{align}$$



                As a result,
                $$a^2 + b^2 + c^2 + d^2 = lambda^2 - 462 ge 214$$



                To see $214$ is the actual minimum, we need to find $(a,b,c,d)$ which satisfies $(*1)$ and $lambda^2 = 676 = 26^2$.



                Flipping all the signs of $a,b,c,d$ if necessary, we can assume $lambda = 26$.



                The third equation $(a+d)(b+c) = 169 = 13^2 = frac14 (26)^2$ tell us
                $a + d = b + c$.

                Introduce $u,v$ such that



                $$(a,b,c,d) = left( frac{13+u}{2}, frac{13+v}{2}, frac{13-v}{2}, frac{13-u}{2}right)$$



                and substitute into the first and second equation and simplify, we obtain



                $$left(frac{u+v}{2}right)^2 = 26quadtext{ and }quad left(frac{u-v}{2}right)^2 = 19$$



                Using this, we find following 4-tuple



                $$(a,b,c,d) = {small left(
                frac{13+sqrt{26}+sqrt{19}}{2},
                frac{13+sqrt{26}-sqrt{19}}{2},
                frac{13-sqrt{26}+sqrt{19}}{2},
                frac{13-sqrt{26}-sqrt{19}}{2}
                right)}$$
                is a solution of the 3 equations in $(*1)$ with $a + b + c + d = 26$.
                One can verify $a^2 + b^2 + c^2 + d^2 = 214$ for this particular solution. As a result, the lower bound $214$ is achievable and $214$ is the smallest possible value we seek.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 27 '18 at 4:26

























                answered Nov 26 '18 at 21:56









                achille hui

                95.6k5130257




                95.6k5130257























                    1














                    Using A.M.-G.M. we get: $$169+143+150=(a+b)(c+d)+(a+c)(b+d)+(a+d)(b+c)=$$



                    $$2ac+2ad+2bc+2bd+2ab+2cd leq 3a^2+3b^2+3c^2+3d^2$$



                    So $$a^2+b^2+c^2+d^2geq 154$$



                    But, unfortenuly this is not the minumum, since variables can not be all the same.






                    share|cite|improve this answer























                    • Cant understand why there's an inquality there
                      – Ahmed I. Elsayed
                      Nov 26 '18 at 21:19










                    • $2xyleq x^2+y^2$........
                      – greedoid
                      Nov 26 '18 at 21:20










                    • 2xy <= x^2 + y^^2 ? not xy <= y^2 + x^2 ?
                      – Ahmed I. Elsayed
                      Nov 26 '18 at 21:24










                    • First.............
                      – greedoid
                      Nov 26 '18 at 21:25










                    • I don't know how your answer addresses the question. You just show a lower bound, not the minimum.
                      – Le Anh Dung
                      Nov 27 '18 at 4:39


















                    1














                    Using A.M.-G.M. we get: $$169+143+150=(a+b)(c+d)+(a+c)(b+d)+(a+d)(b+c)=$$



                    $$2ac+2ad+2bc+2bd+2ab+2cd leq 3a^2+3b^2+3c^2+3d^2$$



                    So $$a^2+b^2+c^2+d^2geq 154$$



                    But, unfortenuly this is not the minumum, since variables can not be all the same.






                    share|cite|improve this answer























                    • Cant understand why there's an inquality there
                      – Ahmed I. Elsayed
                      Nov 26 '18 at 21:19










                    • $2xyleq x^2+y^2$........
                      – greedoid
                      Nov 26 '18 at 21:20










                    • 2xy <= x^2 + y^^2 ? not xy <= y^2 + x^2 ?
                      – Ahmed I. Elsayed
                      Nov 26 '18 at 21:24










                    • First.............
                      – greedoid
                      Nov 26 '18 at 21:25










                    • I don't know how your answer addresses the question. You just show a lower bound, not the minimum.
                      – Le Anh Dung
                      Nov 27 '18 at 4:39
















                    1












                    1








                    1






                    Using A.M.-G.M. we get: $$169+143+150=(a+b)(c+d)+(a+c)(b+d)+(a+d)(b+c)=$$



                    $$2ac+2ad+2bc+2bd+2ab+2cd leq 3a^2+3b^2+3c^2+3d^2$$



                    So $$a^2+b^2+c^2+d^2geq 154$$



                    But, unfortenuly this is not the minumum, since variables can not be all the same.






                    share|cite|improve this answer














                    Using A.M.-G.M. we get: $$169+143+150=(a+b)(c+d)+(a+c)(b+d)+(a+d)(b+c)=$$



                    $$2ac+2ad+2bc+2bd+2ab+2cd leq 3a^2+3b^2+3c^2+3d^2$$



                    So $$a^2+b^2+c^2+d^2geq 154$$



                    But, unfortenuly this is not the minumum, since variables can not be all the same.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 26 '18 at 20:30

























                    answered Nov 26 '18 at 20:23









                    greedoid

                    38.2k114797




                    38.2k114797












                    • Cant understand why there's an inquality there
                      – Ahmed I. Elsayed
                      Nov 26 '18 at 21:19










                    • $2xyleq x^2+y^2$........
                      – greedoid
                      Nov 26 '18 at 21:20










                    • 2xy <= x^2 + y^^2 ? not xy <= y^2 + x^2 ?
                      – Ahmed I. Elsayed
                      Nov 26 '18 at 21:24










                    • First.............
                      – greedoid
                      Nov 26 '18 at 21:25










                    • I don't know how your answer addresses the question. You just show a lower bound, not the minimum.
                      – Le Anh Dung
                      Nov 27 '18 at 4:39




















                    • Cant understand why there's an inquality there
                      – Ahmed I. Elsayed
                      Nov 26 '18 at 21:19










                    • $2xyleq x^2+y^2$........
                      – greedoid
                      Nov 26 '18 at 21:20










                    • 2xy <= x^2 + y^^2 ? not xy <= y^2 + x^2 ?
                      – Ahmed I. Elsayed
                      Nov 26 '18 at 21:24










                    • First.............
                      – greedoid
                      Nov 26 '18 at 21:25










                    • I don't know how your answer addresses the question. You just show a lower bound, not the minimum.
                      – Le Anh Dung
                      Nov 27 '18 at 4:39


















                    Cant understand why there's an inquality there
                    – Ahmed I. Elsayed
                    Nov 26 '18 at 21:19




                    Cant understand why there's an inquality there
                    – Ahmed I. Elsayed
                    Nov 26 '18 at 21:19












                    $2xyleq x^2+y^2$........
                    – greedoid
                    Nov 26 '18 at 21:20




                    $2xyleq x^2+y^2$........
                    – greedoid
                    Nov 26 '18 at 21:20












                    2xy <= x^2 + y^^2 ? not xy <= y^2 + x^2 ?
                    – Ahmed I. Elsayed
                    Nov 26 '18 at 21:24




                    2xy <= x^2 + y^^2 ? not xy <= y^2 + x^2 ?
                    – Ahmed I. Elsayed
                    Nov 26 '18 at 21:24












                    First.............
                    – greedoid
                    Nov 26 '18 at 21:25




                    First.............
                    – greedoid
                    Nov 26 '18 at 21:25












                    I don't know how your answer addresses the question. You just show a lower bound, not the minimum.
                    – Le Anh Dung
                    Nov 27 '18 at 4:39






                    I don't know how your answer addresses the question. You just show a lower bound, not the minimum.
                    – Le Anh Dung
                    Nov 27 '18 at 4:39




















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