<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] To begin with, we note that equality occurs when $a=b=1/2$. Thus, assuming WLOG $a>b$ and using $b=1-a$, let us  differentiate the LHS to show it is increasing: $$2\left(a+\frac{2}{a}\right)\left(1-\frac{2}{a^2}\right)+2\left(1-a+\frac{2}{1-a}\right)\left(-1+\frac{2}{(1-a)^2}\right)>0 $$ $$ \left(a+\frac{2}{a}\right)\left(1-\frac{2}{a^2}\right)>\left(1-a+\frac{2}{(1-a)}\right)\left(1-\frac{2}{(1-a)^2}\right).\tag{1}$$ Again, we differentiate to show the LHS of $(1)$ is increasing: $$\left(1-\frac{2}{a^2}\right)^2+\frac{4}{a^3}\left(a+\frac{2}{a}\right)>0$$ whereas its RHS is decreasing: $$-\left(1-\frac{2}{(1-a)^2}\right)^2-\frac{4}{(1-a)^3}\left(1-a+\frac{2}{1-a}\right)<0,$$ and we are done.
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