<s>[INST] Inequality with x,y,z fractions $\frac{x}{y}+\frac{y}{z+x}+\frac{z}{x}\ge 2$ If $x,y,z>0$, show:
$$\frac{x}{y}+\frac{y}{z+x}+\frac{z}{x}\ge 2$$
I expand and to prove
$$x^3 - 2 x^2 y + x^2 z + x y^2 - x y z + y z^2\ge 0$$
I don't know how to do this.
 [/INST] If $x, y, z > 0$, then we see that
$$
\begin{align}
\frac{x}{y} + \frac{ y }{z+x } + \frac{z }{x } - 2 &= \frac{ x^2(z+x) + xy^2 + yz(z+x) - 2xy(z+x)  }{ xy(z+x) } \\
&= \frac{ yz^2 + x(x - y)z + x\left(x^2 - y^2 \right) }{ xy(z+x) }.
\end{align}
$$
Now consider the numerator, which is a quadratic in $z$; it has the discriminant 
$$ 
\begin{align}
x^2(x- y)^2 - 4 xy \left(x^2 - y^2 \right) &= x^4 - 2x^3y + x^2y^2 - 4x^3y + 4xy^3 \\
&= x^4 - 6x^3y + x^2y^2 + 4xy^3 \\
&= 
\end{align}
$$
The idea is to see if the discriminant is negative for all values of $x, y > 0$, or not.
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