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.
 A: The function $f: x \mapsto (x+\frac2x)^2$ is convex on $(0,\infty)$, so by Jensen's inequality we have
$$
\frac{\left(a+\frac2a\right)^2+\left(b+\frac2b\right)^2}{2} = \frac{f(a)+f(b)}{2} \geq f\left( \frac{a+b}{2} \right) = f\left(\frac12\right) = \left( \frac92 \right)^2 = \frac{81}{4}.
$$
A: 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.
A: your calculations are ok, we are looking at the inequality $$(1-2c)(c^2+4)\geq \frac{65}{2}c^2$$ with $c=ab$ for $c$ we get $$a^2+b^2=1-2ab\geq 2ab$$ from here we get $$\frac{1}{4}\geq ab=c$$ and our inequality is equivalent to $$\frac{-1}{2}(4c-1)(c^2+16c+8)\geq 0$$
ready.
