gt$\Rightarrow$ $\frac{1}{yz}$+$\frac{1}{xz}$+$\frac{1}{xy}$=1
Đặt a=$\frac{1}{x}$;b=$\frac{1}{y}$;c=$\frac{1}{z}$$\Rightarrow$ ab+bc+ca=1
Xét VT=$\frac{2a}{\sqrt{1+a^{2}}}$+$\frac{b}{\sqrt{1+b^{2}}}$+$\frac{c}{\sqrt{1+c^{2}}}$
AD BĐT Cauchy ta được:
$\frac{2a}{\sqrt{1+a^{2}}}$=2a($\sqrt{\frac{1}{a+b}}$.$\sqrt{\frac{1}{a+c}}$)$\leq$2a.$\frac{\frac{1}{a+b}+\frac{1}{a+c}}{2}$=$\frac{a}{a+b}$+$\frac{a}{a+c}$(1)
$\frac{b}{\sqrt{1+b^{2}}}$=2b($\sqrt{\frac{1}{b+a}}$ . $\sqrt{\frac{1}{4(b+c)}}$)$\leq$$\frac{b}{b+a}$+$\frac{b}{4(b+c)}$(2)
tương tự:$\frac{c}{\sqrt{1+c^{2}}}$$\leq$$\frac{c}{c+a}$+$\frac{c}{4(c+b)}$(3)
cộng theo vế của (1)(2)(3) $\Rightarrow$đpcm
dấu''='' xảy ra $\Leftrightarrow$a=$\frac{7}{\sqrt{15}}$;b=c=$\frac{1}{\sqrt{15}}$