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
 A: 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 $\frac{a^+b^2}{a^2b^2}\geq \frac{1}{2}\cdot 16=8$
A: For your revised question, another way is to note that $(x + \frac1x)^2$ is convex, so by Jensen's inequality:
$$\left(a + \frac1a\right)^2 + \left(b + \frac1b\right)^2 \ge 2\left(\frac{a+b}2 + \frac2{a+b}\right)^2=2\left(\frac12 + 2\right)^2=\frac{25}2$$
A: Without loss of generality we can choose $a=sin^2x$ and $b=cos^2x$  for  $x \in \left(0 \: \frac{\pi}{2}\right)$
Now $$a^2+b^2=sin^4x+cos^4x=1-2sin^2xcos^2x=1-\frac{sin^22x}{2}=\frac{3+cos4x}{4} \ge \frac{1}{2}$$ with Equality at $x=\frac{\pi}{4}$
Also $$2sinxcosx \le 1$$ $\implies$
$$\frac{1}{sin^4xcos^4x} \ge 16$$
Now by $CS$ Inequality $$\left(\frac{1}{a^2}+\frac{1}{b^2}\right)(1^2+1^2) \ge\left(\frac{1}{a}+\frac{1}{b}\right)^2=(sec^2x+cosec^2x)^2=(sec^2xcosec^2x)^2=sec^4xcosec^4x=\frac{1}{sin^4xcos^4x} \ge 16$$ Thus
$$\left(\frac{1}{a^2}+\frac{1}{b^2}\right)(1^2+1^2) \ge 16$$
$$\left(\frac{1}{a^2}+\frac{1}{b^2}\right) \ge 8$$
Thus $$\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2 \ge \frac{1}{2}+8+4=\frac{25}{2}$$
