Chứng minh rằng:  $\frac{1}{a^2}+\frac{1}{b^2}+2\ge \frac{8}{a+b}$

Ta có:

  $\begin{array}{l}\frac{1}{a^2}+\frac{1}{b^2}\ge 2\cdot \frac{1}{a}\cdot \frac{1}{b}\\ \iff 2\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\ge \frac{1}{a^2}+\frac{1}{b^2}+2\cdot \frac{1}{a}\cdot \frac{1}{b}\\ \iff 2\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\ge {\left(\frac{1}{a}+\frac{1}{b}\right)}^2\left(1\right)\end{array}$

Mặt khác:

$\begin{array}{l}{\left(\frac{1}{a}+\frac{1}{b}-2\right)}^2\ge 0\\ \iff {\left(\frac{1}{a}+\frac{1}{b}\right)}^2-4\left(\frac{1}{a}+\frac{1}{b}\right)+4\ge 0\\ \iff {\left(\frac{1}{a}+\frac{1}{b}\right)}^2+4\ge 4\left(\frac{1}{a}+\frac{1}{b}\right)\left(2\right)\end{array}$

Lại có: (a + b)² ≥ 4ab

Từ (1), (2), (3) ⇒  $2\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+4\ge {\left(\frac{1}{a}+\frac{1}{b}\right)}^2+4\ge 4\left(\frac{1}{a}+\frac{1}{b}\right)\ge \frac{16}{a+b}$

 $\iff \frac{1}{a^2}+\frac{1}{b^2}+2\ge \frac{8}{a+b}$ (với a, b > 0)