Áp dụng BĐT Bunhiacopxki :
\(\left(x^2+\frac{1}{x^2}\right)\left(1^2+9^2\right)\ge\left(x+\frac{9}{x}\right)^2\)
\(\Leftrightarrow\sqrt{82}\cdot\sqrt{x^2+\frac{1}{x^2}}\ge x+\frac{9}{x}\)
Chứng minh tương tự :
\(\sqrt{82}\cdot\sqrt{y^2+\frac{1}{y^2}}\ge y+\frac{9}{y}\)
\(\sqrt{82}\cdot\sqrt{z^2+\frac{1}{z^2}}\ge z+\frac{9}{z}\)
Cộng theo vế các BĐT :
\(\sqrt{82}\cdot\left(\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}\right)\ge x+y+z+\frac{9}{x}+\frac{9}{y}+\frac{9}{z}\)
Lại có :
\(x+y+z+\frac{9}{x}+\frac{9}{y}+\frac{9}{z}=\frac{9}{x}+81x+\frac{9}{y}+81y+\frac{9}{z}+81z-80\cdot\left(x+y+z\right)\)
\(\ge2\sqrt{\frac{81x\cdot9}{x}}+2\sqrt{\frac{81y\cdot9}{y}}+2\sqrt{\frac{81z\cdot9}{z}}-80\cdot1=82\)
Do đó ta có \(\sqrt{82}\cdot A\ge82\)
\(\Leftrightarrow A\ge\sqrt{82}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\frac{1}{3}\)
