<s>[INST] Sum of $(a+\frac{1}{a})^2$ and $(b+\frac{1}{b})^2$ Prove that:
$$
\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2\ge\frac{25}{2}
$$
if $a,b$ are positive real numbers such that $a+b=1$.
I have tried expanding the squares and rewriting them  such that $a+b$ is a term/part of a term but what I get is completely contradictory to what is asked to prove
 [/INST] For $E=(a+1/a)^2+(b+1/b)^2=a^2+b^2+1/a^2+1/b^2+4$ you have $1=(a+b)^2=a^2+b^2+2ab\leq 2(a^2+b^2)$, so $a^2+b^2\geq 1/2$. Moreover, $\frac{a+b}{2}\geq 2\sqrt{ab}$ so $\frac{1}{(ab)^2}\geq 16$. This implies $$E=a^2+b^2+\frac{a^2+b^2}{a^2b^2}+4\geq 9/2+8=\frac{25}{2},$$
because $\frac{a^+b^2}{a^2b^2}\geq \frac{1}{2}\cdot 16=8$
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