Cho [tex]x,y,z>0;x+y+z\leq \frac{3}{2}[/tex]. Chứng minh rằng [tex]\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}\geq \frac{3}{2}\sqrt{17}[/tex]
Đặt VT=A
Ta có [tex]\frac{\sqrt{17}}{2}\sqrt{x^{2}+\frac{1}{x^2}}=\sqrt{(\frac{1}{4}+4)(x^2+\frac{1}{x^2})}\geq \frac{x}{2}+\frac{2}{x}[/tex];
[tex]\frac{x}{2}+\frac{2}{x}= \frac{x}{2}+\frac{1}{8x}+\frac{15}{8x}\geq \frac{1}{2}+\frac{15}{8x}[/tex]
[tex]\Rightarrow \frac{\sqrt{17}}{2}A\geq \frac{3}{2}+\frac{15}{8}(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq \frac{3}{2}+\frac{15.9}{8(x+y+z)}\geq \frac{3}{2}+\frac{45}{4}=\frac{51}{4}[/tex]
[tex]\Rightarrow A\geq \frac{3\sqrt{17}}{2}[/tex]
Dấu bằng xảy ra [tex]\Leftrightarrow x=y=z=\frac{1}{2}[/tex]