Is right the inequality: $|a|+|b| leq 2|a+ib|$
$begingroup$
I want know if is right this inequality: $forall a,b in mathbb{R}$,
$$|a|+|b| leq 2|a+ib|$$
inequality
edited Dec 10 '18 at 18:22
greedoid
43.3k1153106
asked Dec 10 '18 at 18:18
Deidson SantosDeidson Santos
111
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add a comment |
$begingroup$
I want know if is right this inequality: $forall a,b in mathbb{R}$,
$$|a|+|b| leq 2|a+ib|$$
inequality
edited Dec 10 '18 at 18:22
greedoid
43.3k1153106
asked Dec 10 '18 at 18:18
Deidson SantosDeidson Santos
111
$endgroup$
add a comment |
$begingroup$
I want know if is right this inequality: $forall a,b in mathbb{R}$,
$$|a|+|b| leq 2|a+ib|$$
inequality
edited Dec 10 '18 at 18:22
greedoid
43.3k1153106
asked Dec 10 '18 at 18:18
Deidson SantosDeidson Santos
111
$endgroup$
I want know if is right this inequality: $forall a,b in mathbb{R}$,
$$|a|+|b| leq 2|a+ib|$$
inequality
inequality
edited Dec 10 '18 at 18:22
greedoid
43.3k1153106
asked Dec 10 '18 at 18:18
Deidson SantosDeidson Santos
111
edited Dec 10 '18 at 18:22
greedoid
43.3k1153106
asked Dec 10 '18 at 18:18
Deidson SantosDeidson Santos
111
edited Dec 10 '18 at 18:22
greedoid
43.3k1153106
edited Dec 10 '18 at 18:22
greedoid
43.3k1153106
edited Dec 10 '18 at 18:22
greedoid
43.3k1153106
43.3k1153106
asked Dec 10 '18 at 18:18
Deidson SantosDeidson Santos
111
asked Dec 10 '18 at 18:18
Deidson SantosDeidson Santos
111
asked Dec 10 '18 at 18:18
Deidson SantosDeidson Santos
111
111
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
For real numbers $a$ and $b$ we have
begin{align*}
&&a^2&le a^2+b^2\
&implies&sqrt{a^2}&lesqrt{a^2+b^2}\
&&|a|&lesqrt{a^2+b^2}
end{align*}
In the same way we can prove $|b|lesqrt{a^2+b^2}$. Since $sqrt{a^2+b^2}=|a+ib|$, the asked inequality follows.
answered Dec 10 '18 at 18:24
Ángel Mario GallegosÁngel Mario Gallegos
18.5k11230
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add a comment |
$begingroup$
If you square both sides (which you can, since both are nonegative) you get: $$a^2+2|ab|+b^2leq 4(a^2+b^2)$$
so $$0leq 3x^2-2xy+3y^2= (x-y)^2+2x^2+2y^2$$
which is true. ($x=|a|$ and $y=|b|$).
answered Dec 10 '18 at 18:21
greedoidgreedoid
43.3k1153106
$endgroup$
add a comment |
$begingroup$
Hint: Use that $$|a+ib|=sqrt{a^2+b^2}$$ and then square it.
answered Dec 10 '18 at 18:22
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.9k42866
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add a comment |
$begingroup$
The numbers of both sides are non-negatives. Since $(|a|+|b|)^2 leq (2|a+ib|)^2$ (by a straightforward computation), we conclude that $$|a|+|b| leq 2|a+ib|.$$
answered Dec 10 '18 at 18:27
user376343user376343
3,7883828
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For real numbers $a$ and $b$ we have
begin{align*}
&&a^2&le a^2+b^2\
&implies&sqrt{a^2}&lesqrt{a^2+b^2}\
&&|a|&lesqrt{a^2+b^2}
end{align*}
In the same way we can prove $|b|lesqrt{a^2+b^2}$. Since $sqrt{a^2+b^2}=|a+ib|$, the asked inequality follows.
answered Dec 10 '18 at 18:24
Ángel Mario GallegosÁngel Mario Gallegos
18.5k11230
$endgroup$
add a comment |
$begingroup$
For real numbers $a$ and $b$ we have
begin{align*}
&&a^2&le a^2+b^2\
&implies&sqrt{a^2}&lesqrt{a^2+b^2}\
&&|a|&lesqrt{a^2+b^2}
end{align*}
In the same way we can prove $|b|lesqrt{a^2+b^2}$. Since $sqrt{a^2+b^2}=|a+ib|$, the asked inequality follows.
answered Dec 10 '18 at 18:24
Ángel Mario GallegosÁngel Mario Gallegos
18.5k11230
$endgroup$
add a comment |
$begingroup$
For real numbers $a$ and $b$ we have
begin{align*}
&&a^2&le a^2+b^2\
&implies&sqrt{a^2}&lesqrt{a^2+b^2}\
&&|a|&lesqrt{a^2+b^2}
end{align*}
In the same way we can prove $|b|lesqrt{a^2+b^2}$. Since $sqrt{a^2+b^2}=|a+ib|$, the asked inequality follows.
answered Dec 10 '18 at 18:24
Ángel Mario GallegosÁngel Mario Gallegos
18.5k11230
$endgroup$
For real numbers $a$ and $b$ we have
begin{align*}
&&a^2&le a^2+b^2\
&implies&sqrt{a^2}&lesqrt{a^2+b^2}\
&&|a|&lesqrt{a^2+b^2}
end{align*}
In the same way we can prove $|b|lesqrt{a^2+b^2}$. Since $sqrt{a^2+b^2}=|a+ib|$, the asked inequality follows.
answered Dec 10 '18 at 18:24
Ángel Mario GallegosÁngel Mario Gallegos
18.5k11230
answered Dec 10 '18 at 18:24
Ángel Mario GallegosÁngel Mario Gallegos
18.5k11230
answered Dec 10 '18 at 18:24
Ángel Mario GallegosÁngel Mario Gallegos
18.5k11230
answered Dec 10 '18 at 18:24
Ángel Mario GallegosÁngel Mario Gallegos
18.5k11230
18.5k11230
add a comment |
add a comment |
$begingroup$
If you square both sides (which you can, since both are nonegative) you get: $$a^2+2|ab|+b^2leq 4(a^2+b^2)$$
so $$0leq 3x^2-2xy+3y^2= (x-y)^2+2x^2+2y^2$$
which is true. ($x=|a|$ and $y=|b|$).
answered Dec 10 '18 at 18:21
greedoidgreedoid
43.3k1153106
$endgroup$
add a comment |
$begingroup$
If you square both sides (which you can, since both are nonegative) you get: $$a^2+2|ab|+b^2leq 4(a^2+b^2)$$
so $$0leq 3x^2-2xy+3y^2= (x-y)^2+2x^2+2y^2$$
which is true. ($x=|a|$ and $y=|b|$).
answered Dec 10 '18 at 18:21
greedoidgreedoid
43.3k1153106
$endgroup$
add a comment |
$begingroup$
If you square both sides (which you can, since both are nonegative) you get: $$a^2+2|ab|+b^2leq 4(a^2+b^2)$$
so $$0leq 3x^2-2xy+3y^2= (x-y)^2+2x^2+2y^2$$
which is true. ($x=|a|$ and $y=|b|$).
answered Dec 10 '18 at 18:21
greedoidgreedoid
43.3k1153106
$endgroup$
If you square both sides (which you can, since both are nonegative) you get: $$a^2+2|ab|+b^2leq 4(a^2+b^2)$$
so $$0leq 3x^2-2xy+3y^2= (x-y)^2+2x^2+2y^2$$
which is true. ($x=|a|$ and $y=|b|$).
answered Dec 10 '18 at 18:21
greedoidgreedoid
43.3k1153106
answered Dec 10 '18 at 18:21
greedoidgreedoid
43.3k1153106
answered Dec 10 '18 at 18:21
greedoidgreedoid
43.3k1153106
answered Dec 10 '18 at 18:21
greedoidgreedoid
43.3k1153106
43.3k1153106
add a comment |
add a comment |
$begingroup$
Hint: Use that $$|a+ib|=sqrt{a^2+b^2}$$ and then square it.
answered Dec 10 '18 at 18:22
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.9k42866
$endgroup$
add a comment |
$begingroup$
Hint: Use that $$|a+ib|=sqrt{a^2+b^2}$$ and then square it.
answered Dec 10 '18 at 18:22
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.9k42866
$endgroup$
add a comment |
$begingroup$
Hint: Use that $$|a+ib|=sqrt{a^2+b^2}$$ and then square it.
answered Dec 10 '18 at 18:22
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.9k42866
$endgroup$
Hint: Use that $$|a+ib|=sqrt{a^2+b^2}$$ and then square it.
answered Dec 10 '18 at 18:22
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.9k42866
answered Dec 10 '18 at 18:22
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.9k42866
answered Dec 10 '18 at 18:22
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.9k42866
answered Dec 10 '18 at 18:22
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.9k42866
75.9k42866
add a comment |
add a comment |
$begingroup$
The numbers of both sides are non-negatives. Since $(|a|+|b|)^2 leq (2|a+ib|)^2$ (by a straightforward computation), we conclude that $$|a|+|b| leq 2|a+ib|.$$
answered Dec 10 '18 at 18:27
user376343user376343
3,7883828
$endgroup$
add a comment |
$begingroup$
The numbers of both sides are non-negatives. Since $(|a|+|b|)^2 leq (2|a+ib|)^2$ (by a straightforward computation), we conclude that $$|a|+|b| leq 2|a+ib|.$$
answered Dec 10 '18 at 18:27
user376343user376343
3,7883828
$endgroup$
add a comment |
$begingroup$
The numbers of both sides are non-negatives. Since $(|a|+|b|)^2 leq (2|a+ib|)^2$ (by a straightforward computation), we conclude that $$|a|+|b| leq 2|a+ib|.$$
answered Dec 10 '18 at 18:27
user376343user376343
3,7883828
$endgroup$
The numbers of both sides are non-negatives. Since $(|a|+|b|)^2 leq (2|a+ib|)^2$ (by a straightforward computation), we conclude that $$|a|+|b| leq 2|a+ib|.$$
answered Dec 10 '18 at 18:27
user376343user376343
3,7883828
answered Dec 10 '18 at 18:27
user376343user376343
3,7883828
answered Dec 10 '18 at 18:27
user376343user376343
3,7883828
answered Dec 10 '18 at 18:27
user376343user376343
3,7883828
3,7883828
add a comment |
add a comment |
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