If I have $a>0$ and $ b >0 $ and $a+b=1$ then how can I find the minimum value of $(a+1/a)^2 + (b+1/b)^2$. I just tried to do it by expanding the expression just because I knew the graph of $x^2+1/x^2 $ and for that I had the minimum value as 2 but since there are two variables $a,b$ in the problem with the condition that both are less than 1 I'm not able to proceed from here. Any help will be appreciated .Thanks in advance.
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I think it should be $$\frac{25}{2}$$ – Dr. Sonnhard Graubner Oct 21 '18 at 18:23
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Yes the answer given is correct . Can you give a hint how you did it? – Monu Monu Oct 21 '18 at 18:25
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https://math.stackexchange.com/questions/487486/proving-inequality-a-frac1a2-b-frac1b2-geq-frac252-for – lab bhattacharjee Oct 21 '18 at 18:30
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Hint: Let $f(x)=\left(x+\frac{1}{x}\right)^2$, then the given expression can be thought of as $g(a)=f(a)+f(1-a)$. Exploit symmetry with $a=1/2$ and use convexity of $f$. – Anurag A Oct 21 '18 at 18:31
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Thanks for help – Monu Monu Oct 21 '18 at 18:41
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Hint: By expanding we get $$a^2+b^2+\frac{1}{a^2}+\frac{1}{b^2}\geq \frac{17}{2}$$
This is $$(a^2+b^2)\left(1+\frac{1}{a^2b^2}\right)\geq \frac{17}{2} $$ Now we have $$a+b=1$$ so $$a^2+b^2=1-2ab$$ So you have to prove that
$$(1-2ab)(1+\frac{1}{a^2b^2})\geq \frac{17}{2}$$ Can you finish?
Dr. Sonnhard Graubner
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This problem is as old as a volkswagen beetle classic ( that I have ,haha ). My answer is problably one of the $20$ different "proofs" that you can find here or google. I try to be a little different if possible...
Using Cauchy-Schwarz inequality twice: $(1^2+1^2)\left(\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2\right)\ge \left(1\cdot\left(a+\dfrac{1}{a}\right)+1\cdot\left(b+\dfrac{1}{b}\right)\right)^2=\left(a+b+\dfrac{1^2}{a}+\dfrac{1^2}{b}\right)^2\ge \left(a+b+\dfrac{(1+1)^2}{a+b}\right)^2=\left(1+\dfrac{2^2}{1}\right)^2= 5^2=25\implies \left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2\ge \dfrac{25}{2}$, with $ =$ occurs when $a = b = \dfrac{1}{2}$ .
DeepSea
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