Question 46
If $x$, $y$, $z$ are positive numbers with $x + y + z = 1$, show that:
$$
\begin{align*}
a)\ 1 - \frac{x^2-yz}{x^2+x} &= \frac{(1-y)(1-z)}{x^2+x}; \\
b)\ \frac{x^2-yz}{x^2+x} + \frac{y^2-zx}{y^2+y} + \frac{z^2-xy}{z^2+z} &\le 0.
\end{align*}
$$
$$
\begin{align*}
a)\ 1 - \frac{x^2-yz}{x^2+x} &= \frac{(1-y)(1-z)}{x^2+x}; \\
b)\ \frac{x^2-yz}{x^2+x} + \frac{y^2-zx}{y^2+y} + \frac{z^2-xy}{z^2+z} &\le 0.
\end{align*}
$$