$AM-GM$ for $(a + \frac{1}{a})^2 + (b + \frac{1}{b})^2 \geqslant \frac{25}{2}$

Question

Let $a, b \in \mathbb{R_+}$ such that, $a + b = 1$. Show that $$(a + \frac{1}{a})^2 + (b + \frac{1}{b})^2 \geqslant \frac{25}{2}.$$

This was on a problem set for practising $AM-GM$ and I couldn’t figure out how to approach this. What would be the optimal way to show this?

Answer 1

Using the two versions of the AM-GM inequality: $X^2+Y^2 \ge 2XY$ for $X = a+\frac{1}{a}, Y = \frac{5}{2}$and again for $X = b+\frac{1}{b}, Y = \frac{5}{2}$, and also $\frac{1}{X}+\frac{1}{Y} \ge \frac{4}{X+Y}$ for $X = a, Y = b$:

$\left(a+\frac{1}{a}\right)^2+\left(\frac{5}{2}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(\frac{5}{2}\right)^2\ge 5\left(a+\frac{1}{a}\right)+5\left(b+\frac{1}{b}\right)= 5(a+b)+5\left(\frac{1}{a}+\frac{1}{b}\right)= 5 + 5\left(\frac{1}{a}+\frac{1}{b}\right)\ge 5+5\cdot\frac{4}{a+b}= 5+20=25\implies \left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2\ge \frac{25}{2}$.

Answer 2

Using $a+b=1$, it is easy to see $ab\le \frac14$. So, $$(a + \frac{1}{a})^2 + (b + \frac{1}{b})^2 \ge \frac12\bigg(a + \frac{1}{a}+b + \frac{1}{b}\bigg)^2=\frac12(1+\frac{1}{ab})^2\ge\frac12(1+4)^2=\frac{25}{2}.$$