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Am trying to solve the inequality $$(a+\frac{1}{a})^2 + (b+\frac{1}{b})^2 \geq 25/2$$ subject to $a,b$ are positive numbers whose sum is $1$.

I used the AM-GM inequality, $\frac{a+b}{2}\geq\sqrt{ab}$. I Know that AM-GM inequality gives a minimum if $a=b$, since that is where equality exists. I reasoned that I could substitute $1/2$ for $a$ and $b$ in the original inequality getting $\left(\frac{1}{2} + \frac{1}{\frac{1}{2}}\right)^2 + \left(\frac{1}{2} + \frac{1}{\frac{1}{2}}\right)^2 \geq (2.5)^2 +(2.5)^2 = 6.25 + 6.25 = 12.5 = 25/2$

Works out but "doesn't feel right." Comments or alternative approaches please.

Anindya Prithvi
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gary
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    Do you know tangent line method? That is one of the ways. – Math Lover Dec 01 '20 at 18:02
  • I just found this link. Please check. It has links to a few others too. https://math.stackexchange.com/questions/487486/proving-inequality-a-frac1a2-b-frac1b2-geq-frac252-for – Math Lover Dec 01 '20 at 18:09

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