Sum of $(a+\frac{1}{a})^2$ and $(b+\frac{1}{b})^2$

Solution 1:

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 $\frac{a^+b^2}{a^2b^2}\geq \frac{1}{2}\cdot 16=8$

Solution 2:

For your revised question, another way is to note that $(x + \frac1x)^2$ is convex, so by Jensen's inequality: $$\left(a + \frac1a\right)^2 + \left(b + \frac1b\right)^2 \ge 2\left(\frac{a+b}2 + \frac2{a+b}\right)^2=2\left(\frac12 + 2\right)^2=\frac{25}2$$