Let $x, y, z >0$ . Prove that:
$$\frac{3xy}{xy+x+y}+ \frac{3yz}{yz+y+z}+ \frac{3zx}{zx+z+x}\leq 2+ \frac{x^2+y^2+z^2}{3}$$
We have succesively:
$$\frac{3xy}{xy+x+y}+ \frac{3yz}{yz+y+z}+ \frac{3zx}{zx+z+x}\leq 2+ \frac{x^2+y^2+z^2}{3}$$
(Câtâlin Cristea)
Solution
\begin{align*}
2+ \frac{x^2+y^2+z^2}{3} &=\sum \frac{x^2+y^2+4}{6} \\
&\geq \sum \sqrt[6]{x^2y^2} \\
&=\sum \frac{xy}{\sqrt[3]{(xy)xy}} \\
&\geq \sum \frac{xy}{\left ( xy+x+y \right )/3}\\
&=\sum \frac{3xy}{xy+x+y}
\end{align*}
2+ \frac{x^2+y^2+z^2}{3} &=\sum \frac{x^2+y^2+4}{6} \\
&\geq \sum \sqrt[6]{x^2y^2} \\
&=\sum \frac{xy}{\sqrt[3]{(xy)xy}} \\
&\geq \sum \frac{xy}{\left ( xy+x+y \right )/3}\\
&=\sum \frac{3xy}{xy+x+y}
\end{align*}
In the first line you need to delete $z^2$ and instead write
ReplyDelete$$\sum\frac{x^2+y^2+4}{6}.$$