Cho 3 số dương a, b, c thỏa mãn: a + b + c + ab + ca + bc = 6abc.

Chứng minh rằng:  $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge 3$.

Từ a + b + c + ab + ca + bc = 6abc ta có:  $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=6$

Ta có:  $\frac{1}{a^2}+1\ge \frac{2}{a};\frac{1}{b^2}+1\ge \frac{2}{b};\frac{1}{c^2}+1\ge \frac{2}{c}$

Và  $\frac{1}{a^2}+\frac{1}{b^2}\ge \frac{2}{ab};\frac{1}{b^2}+\frac{1}{c^2}\ge \frac{2}{bc};\frac{1}{a^2}+\frac{1}{c^2}\ge \frac{2}{ac}$

Cộng các vế các BĐT trên ta có:

$\begin{array}{l}3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+1\right)\ge 2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\\ \iff 3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+1\right)\ge 12\\ \iff \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+1\ge 4\\ \iff \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge 3\end{array}$

Dấu “ = ” xảy ra khi a = b = c = 1