Inequality with roots

Let $x, y, z$ be positive real number such that $x+y+z=1$. Prove that:

$$\sqrt{\frac{xy}{xy+z}}+ \sqrt{\frac{yz}{yz+x}}+ \sqrt{\frac{zx}{zx+y}} \leq \frac{3}{2}$$

Solution

We note that:

\begin{align*}
\sqrt{\frac{xy}{xy+z}} &=\sqrt{\frac{xy}{xy+z \left ( x+y+z \right )}} \\
 &= \sqrt{\frac{xy}{(x+z) \left ( y+z \right )}}\\
 & \leq \frac{1}{2}\left ( \frac{x}{x+z}+ \frac{y}{y+z} \right )
\end{align*}

Similarly we have that:

$$\sqrt{\frac{yz}{yz+z}} \leq \frac{1}{2}\left ( \frac{y}{y+x}+ \frac{z}{z+x} \right ) \quad , \quad \sqrt{\frac{zx}{zx+y}} \leq \frac{1}{2}\left ( \frac{x}{x+y}+ \frac{z}{z+y} \right )$$

Adding cyclic we have that:

\begin{align*}
\sum \sqrt{\frac{xy}{xy+z}} &\leq \frac{1}{2}\sum \left ( \frac{x}{x+z} + \frac{y}{y+z} \right ) \\
 &= \frac{3}{2}
\end{align*}

proving the result.