Smallest value of $a^2 + b^2 + c^2+ d^2$, given values for $(a+b)(c+d)$, $(a+c)(b+d)$, and $(a+d)(b+c)$
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
edited Nov 27 '18 at 4:46
Blue
47.7k870151
asked Nov 26 '18 at 19:20
Ahmed I. Elsayed
1134
add a comment |
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
edited Nov 27 '18 at 4:46
Blue
47.7k870151
asked Nov 26 '18 at 19:20
Ahmed I. Elsayed
1134
add a comment |
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
edited Nov 27 '18 at 4:46
Blue
47.7k870151
asked Nov 26 '18 at 19:20
Ahmed I. Elsayed
1134
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
edited Nov 27 '18 at 4:46
Blue
47.7k870151
asked Nov 26 '18 at 19:20
Ahmed I. Elsayed
1134
edited Nov 27 '18 at 4:46
Blue
47.7k870151
asked Nov 26 '18 at 19:20
Ahmed I. Elsayed
1134
edited Nov 27 '18 at 4:46
Blue
47.7k870151
edited Nov 27 '18 at 4:46
Blue
47.7k870151
edited Nov 27 '18 at 4:46
Blue
47.7k870151
47.7k870151
asked Nov 26 '18 at 19:20
Ahmed I. Elsayed
1134
asked Nov 26 '18 at 19:20
Ahmed I. Elsayed
1134
asked Nov 26 '18 at 19:20
Ahmed I. Elsayed
1134
1134
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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.
answered Nov 26 '18 at 21:56
achille hui
95.6k5130257
add a comment |
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.
answered Nov 26 '18 at 20:23
greedoid
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
add a comment |
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2 Answers
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered Nov 26 '18 at 21:56
achille hui
95.6k5130257
add a comment |
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.
answered Nov 26 '18 at 21:56
achille hui
95.6k5130257
add a comment |
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.
answered Nov 26 '18 at 21:56
achille hui
95.6k5130257
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.
answered Nov 26 '18 at 21:56
achille hui
95.6k5130257
edited Nov 27 '18 at 4:26
answered Nov 26 '18 at 21:56
achille hui
95.6k5130257
answered Nov 26 '18 at 21:56
achille hui
95.6k5130257
answered Nov 26 '18 at 21:56
achille hui
95.6k5130257
95.6k5130257
add a comment |
add a comment |
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.
answered Nov 26 '18 at 20:23
greedoid
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
add a comment |
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.
answered Nov 26 '18 at 20:23
greedoid
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
add a comment |
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.
answered Nov 26 '18 at 20:23
greedoid
38.2k114797
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.
answered Nov 26 '18 at 20:23
greedoid
38.2k114797
edited Nov 26 '18 at 20:30
answered Nov 26 '18 at 20:23
greedoid
38.2k114797
answered Nov 26 '18 at 20:23
greedoid
38.2k114797
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
add a comment |
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
add a comment |
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