Đặt \(\left(\sqrt{x};\sqrt{y};\sqrt{z}\right)=\left(a;b;c\right)\)
BĐT cần chứng minh: \(\frac{a+b}{c^2}+\frac{b+c}{a^2}+\frac{c+a}{b^2}\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(VT=a\left(\frac{1}{b^2}+\frac{1}{c^2}\right)+b\left(\frac{1}{a^2}+\frac{1}{c^2}\right)+c\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\ge2\left(\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}\right)\)
Mà: \(\frac{a}{bc}+\frac{c}{ab}\ge\frac{2}{b}\) ; \(\frac{a}{bc}+\frac{b}{ac}\ge\frac{2}{c}\) ; \(\frac{c}{ab}+\frac{b}{ac}\ge\frac{2}{a}\)
\(\Rightarrow2\left(\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}\right)\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow VT\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\) (đpcm)
