$x+y+z=xyz \leftrightarrow \dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}=1$
Đặt $\dfrac{1}{x}=a \\ \dfrac{1}{y}=b \\ \dfrac{1}{z}=c$
$\rightarrow ab+bc+ca=1 $
Ta có: $\dfrac{2}{\sqrt{1+x^2}}=\dfrac{2}{\sqrt{1+\dfrac{1}{a^2}}}=\dfrac{2a}
{\sqrt{a^2+1}}=\dfrac{2a}{\sqrt{a^2+ab+bc+ca}}= \dfrac{ 2a }{ \sqrt{(a+b)(a+c)} }
(1)$
Tương tự $\dfrac{1}{\sqrt{1+y^2}}=\dfrac{b}{\sqrt{(b+c)(b+a)}}(2) \\ \dfrac{1}
{\sqrt{1+z^2}}=\dfrac{c}{\sqrt{(c+a)(c+b)}}(3)$
Lại có $\dfrac{2a}{\sqrt{(a+b)(a+c)}} \leq \dfrac{a}{a+b}+\dfrac{a}{a+c}(4)$
$\dfrac{b}{\sqrt{(b+c)(b+a)}}=\dfrac{2b}{\sqrt{(4b+4c)(b+a)}} \leq \dfrac{b}
{4b+4c}+\dfrac{b}{a+b}(5)$
$\dfrac{c}{\sqrt{(b+c)(c+a)}}=\dfrac{2c}{\sqrt{(4b+4c)(c+a)}} \leq \dfrac{c}
{4b+4c}+\dfrac{c}{c+a}(6)$
Từ $(1);(2);(3);(4);(5);(6)$ ta có:
$\dfrac{2}{\sqrt{1+x^2}}+\dfrac{1}{\sqrt{1+y^2}}+ \dfrac{ 1 }{ \sqrt{1+z^2} } \leq
\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{4b+4c}+ \dfrac{ b }{ a+b }+\dfrac{c}
{4b+4c}+\dfrac{c}{c+a}=\dfrac{a+b}{a+b}+\dfrac{a+c}{a+c}+\dfrac{b+c}
{4b+4c}=\dfrac{9}{4}$
Dấu $"="$ xảy ra $\leftrightarrow x=\dfrac{\sqrt{15}}{7} \ \ \ \
y=z=\sqrt{15}$
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