[tex]\frac{1}{\sqrt[4]{x^3+2y^3+6}}\leq \frac{1}{\sqrt[4]{xy(x+y)+3y+4}}=\frac{1}{\sqrt[4]{\frac{x+y+3yz+4z}{z}}}\leq \frac{1}{\sqrt[4]{\frac{3xyz+3yz+3z}{z}}}=\frac{1}{\sqrt[4]{3(xy+y+1)}}[/tex]
làm tượng tự nha :
[tex]xy+1+y=xyz(1+\frac{1}{z}+\frac{1}{xz})=y(xz+x+1)[/tex]
[tex]xy+1+y=xy(1+z+yz)[/tex]
[tex]\frac{1}{\sqrt[4]{x^3+2y^3+6}}+\frac{1}{\sqrt[4]{y^3+2z^3+6}}+\frac{1}{\sqrt[4]{z^3+2x^3+6}}\leq \frac{1}{\sqrt[4]{3(xy+y+1)}}\left ( \sqrt[4]{xy}+\sqrt[4]{y}+1 \right )[/tex]
[tex](xy+y+1)\geq \frac{(\sqrt{xy}+\sqrt{y}+1)^2}{3}\geq \frac{(\sqrt[4]{xy}+\sqrt[4]{y}+1)^4}{27}[/tex]
[tex]\frac{1}{\sqrt[4]{3(xy+y+1)}}\left ( \sqrt[4]{xy}+\sqrt[4]{y}+1 \right )\leq \frac{1}{\sqrt[4]{\frac{(\sqrt[4]{xy}+\sqrt[4]{y}+1)^4}{9}}}(\sqrt[4]{xy}+\sqrt[4]{y}+1)=\sqrt{3}[/tex]
dấu "=" khi x=y=z=1
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