<s>[INST] Proving $\left(a+\frac{2}{a}\right)^2+\left(b+\frac{2}{b}\right)^2\ge \frac{81}{2}$ for all positive real $a,b$ such that $a+b=1$ I approached this problem in two different ways, but only one was successful. I'll post the latter as an answer, while here follows the first approach:
I expanded the squares:
$$a^2+\frac{4}{a^2}+4+b^2+\frac{4}{b^2}+4\ge\frac{81}{2} \\ a^2+b^2+\frac{4}{a^2}+\frac{4}{b^2}\ge \frac{65}{2}$$ and then multiplied both sides by $a^2b^2$ to get $$a^2b^2(a^2+b^2)+4(a^2+b^2)\ge\frac{65}{2}a^2b^2 \\ (a^2+b^2)(a^2b^2+4)\ge\frac{65}{2}a^2b^2, $$ which, combining $a+b=1$ and $(a+b)^2=a^2+b^2+2ab$, can be rewritten as $$(1-2ab)(a^2b^2+4)\ge\frac{65}{2}a^2b^2.\tag{$\star$}$$ Setting $c=ab$ makes $(\star)$ an inequality in one variable, but it didn't help me. Is this totally the wrong track? And, I was wondering if there are other approaches besides what I came up with.
 [/INST] The function $f: x \mapsto (x+\frac2x)^2$ is convex on $(0,\infty)$, so by Jensen's inequality we have
$$
\frac{\left(a+\frac2a\right)^2+\left(b+\frac2b\right)^2}{2} = \frac{f(a)+f(b)}{2} \geq f\left( \frac{a+b}{2} \right) = f\left(\frac12\right) = \left( \frac92 \right)^2 = \frac{81}{4}.
$$
</s>