You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser.
Đặt [tex]\frac{1}{xy}=a;\frac{1}{yz}=b;\frac{1}{zx}=c\Rightarrow a+b+c=1[/tex]
Có $Q=\sum \frac{x}{\sqrt{yz(1+x^{2})}}$
$=\sum \frac{\frac{x}{xyz}}{\sqrt{\frac{yz(1+x^{2})}{x^{2}y^{2}z^{2}}}}$
$=\sum \frac{\frac{1}{yz}}{\sqrt{\frac{1}{yz}.(\frac{1}{x^{2}}+1)}}$
$=\sum \frac{b}{\sqrt{b.(\frac{ac}{b}+1)}}$
$=\sum \frac{b}{\sqrt{ac+b}}$
$=\sum \frac{b}{\sqrt{ac+b(a+b+c)}}$ ( vì a+b+c=1)
$=\sum \frac{b}{\sqrt{(b+a)(b+c)}}$
$\leq \frac{1}{2}.\sum [b.(\frac{1}{b+a}+\frac{1}{b+c})]$ ( BĐT Cauchy ngược)
$=\frac{1}{2}.\sum (\frac{b}{b+a}+\frac{b}{b+c})$
$=\frac{1}{2}.3=\frac{3}{2}$
Dấu "=" xảy ra <=> [tex]\Leftrightarrow a=b=c=\frac{1}{3}\Leftrightarrow x=y=z=\sqrt{3}[/tex]