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Prove that:

$$ \left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2\ge\frac{25}{2} $$

if $a,b$ are positive real numbers such that $a+b=1$.

I have tried expanding the squares and rewriting them such that $a+b$ is a term/part of a term but what I get is completely contradictory to what is asked to prove

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  • $\begingroup$ Maybe use the fact that $(a+b)^2=1$ also. $\endgroup$ – Sam Weatherhog Nov 3 '15 at 1:43
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    $\begingroup$ Is that a typo, or is it not symmetric? That is, did you intend the $\color{Red}2$ to be in $(a+\frac1{a^{\color{Red}2}})^2$ and not in $(b+\frac1b)^2$? $\endgroup$ – Akiva Weinberger Nov 3 '15 at 1:47
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    $\begingroup$ Should the second term be $\left(b+\frac{1}{b^2}\right)^2$? $\endgroup$ – Sam Weatherhog Nov 3 '15 at 1:47
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    $\begingroup$ i'm guessing that the first expression should be $(a+\frac 1a)^2$ $\endgroup$ – WW1 Nov 3 '15 at 1:49
  • $\begingroup$ I'm sorry. Its (a+1/a)^2 $\endgroup$ – Gayatri Nov 3 '15 at 11:44
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For $E=(a+1/a)^2+(b+1/b)^2=a^2+b^2+1/a^2+1/b^2+4$ you have $1=(a+b)^2=a^2+b^2+2ab\leq 2(a^2+b^2)$, so $a^2+b^2\geq 1/2$. Moreover, $\frac{a+b}{2}\geq 2\sqrt{ab}$ so $\frac{1}{(ab)^2}\geq 16$. This implies $$E=a^2+b^2+\frac{a^2+b^2}{a^2b^2}+4\geq 9/2+8=\frac{25}{2},$$

because $\frac{a^+b^2}{a^2b^2}\geq \frac{1}{2}\cdot 16=8$

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  • $\begingroup$ Very cool solution. Much more direct than my method. $\endgroup$ – goblin Nov 3 '15 at 2:22
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    $\begingroup$ Nice +1. For the current version of the problem, you may say $\dfrac1{a^2}>\dfrac1a$, so there is strict inequality. $\endgroup$ – Macavity Nov 3 '15 at 2:28
  • $\begingroup$ Really elegant solution! +1 $\endgroup$ – ZFR Nov 3 '15 at 11:55
  • $\begingroup$ Thank you for the solutions provided. I am tyring to understand it. But I am not able to understand this part. How was this derived. :$\frac{a+b}{2}$ is greater than oe equal to $2*\sqrt{ab}$ $\endgroup$ – Gayatri Nov 3 '15 at 13:44
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    $\begingroup$ It is a typo, it should be $a+b\ge 2\sqrt{ab}$, which is really what is used in the next line. This follows from $(\sqrt a - \sqrt b)^2\ge 0$ and is well known, very useful to remember. $\endgroup$ – Macavity Nov 3 '15 at 14:14
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For your revised question, another way is to note that $(x + \frac1x)^2$ is convex, so by Jensen's inequality: $$\left(a + \frac1a\right)^2 + \left(b + \frac1b\right)^2 \ge 2\left(\frac{a+b}2 + \frac2{a+b}\right)^2=2\left(\frac12 + 2\right)^2=\frac{25}2$$

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Without loss of generality we can choose $a=sin^2x$ and $b=cos^2x$ for $x \in \left(0 \: \frac{\pi}{2}\right)$

Now $$a^2+b^2=sin^4x+cos^4x=1-2sin^2xcos^2x=1-\frac{sin^22x}{2}=\frac{3+cos4x}{4} \ge \frac{1}{2}$$ with Equality at $x=\frac{\pi}{4}$

Also $$2sinxcosx \le 1$$ $\implies$

$$\frac{1}{sin^4xcos^4x} \ge 16$$

Now by $CS$ Inequality $$\left(\frac{1}{a^2}+\frac{1}{b^2}\right)(1^2+1^2) \ge\left(\frac{1}{a}+\frac{1}{b}\right)^2=(sec^2x+cosec^2x)^2=(sec^2xcosec^2x)^2=sec^4xcosec^4x=\frac{1}{sin^4xcos^4x} \ge 16$$ Thus

$$\left(\frac{1}{a^2}+\frac{1}{b^2}\right)(1^2+1^2) \ge 16$$

$$\left(\frac{1}{a^2}+\frac{1}{b^2}\right) \ge 8$$

Thus $$\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2 \ge \frac{1}{2}+8+4=\frac{25}{2}$$

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