Chỉ đúng với a;b;c dương:
[tex]T=\frac{x}{1-yz}+\frac{y}{1-zx}+\frac{z}{1-xy}\\\rightarrow \frac{T^2}{3}=\frac{(\frac{x}{1-yz}+\frac{y}{1-zx}+\frac{z}{1-xy})^2}{3}\leq \left( {\frac{x}{{1 - yz}}} \right)^2 + \left( {\frac{y}{{1 - zx}}} \right)^2 + \left( {\frac{z}{{1 - xy}}} \right)^2[/tex] [tex]T=\frac{x}{1-yz}+\frac{y}{1-zx}+\frac{z}{1-xy}\\\rightarrow \frac{T^2}{3}=\frac{(\frac{x}{1-yz}+\frac{y}{1-zx}+\frac{z}{1-xy})^2}{3}\leq \left( {\frac{x}{{1 - yz}}} \right)^2 + \left( {\frac{y}{{1 - zx}}} \right)^2 + \left( {\frac{z}{{1 - xy}}} \right)^2[/tex] ( Schwarz ngược dấu nhé)
Lại có:
[tex]\left( {\frac{x}{{1 - yz}}} \right)^2 + \left( {\frac{y}{{1 - zx}}} \right)^2 + \left( {\frac{z}{{1 - xy}}} \right)^2\leq \dfrac{x^2}{\left(1- \frac{y^2+z^2}{2}\right)^2}+ \dfrac{y^2}{\left(1- \frac{x^2+z^2}{2}\right)^2}+ \dfrac{z^2}{\left(1- \frac{x^2+y^2}{2}\right)^2}= \dfrac{4x^2}{(1+x^2)^2}+ \dfrac{4y^2}{(1+y^2)^2}+ \dfrac{4x^2}{(1+x^2)^2}=\dfrac{4x^2}{(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+x^2)^2}+ \dfrac{4y^2}{(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+y^2)^2}+ \dfrac{4x^2}{(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+x^2)^2}\le \sum \frac{4x^2}{16.\sqrt[4]{(\frac{1}{3^3}.x^2)^2}}= \dfrac{9}{4\sqrt{3}}(x+y+z)\le \dfrac{9}{4\sqrt{3}}\cdot \sqrt{3(x^2+y^2+z^2)}= \dfrac{9}{4}[/tex]
Vậy
[tex]\frac{T^2}{3}\leq \frac{9}{4}\rightarrow T^2\leq \frac{27}{4}\rightarrow -\frac{3\sqrt{3}}{2}\leq T\leq \frac{3\sqrt{3}}{2}[/tex]
Dấu [tex]"="\Leftrightarrow x=y=z= \dfrac{1}{\sqrt{3}}[/tex]
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