Inequality with x,y,z fractions $\frac{x}{y}+\frac{y}{z+x}+\frac{z}{x}\ge 2$

Question

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.

Answer 1

This might be pretty cumbersome to prove, if you don't see the trick is to sum $1$ in both sides:

$$\frac{x}{y}+\frac{y}{z+x}+\frac{z}{x} + 1\geq 3$$

And this follows directly from AM-GM:

$$\frac{x}{y}+\frac{y}{z+x}+\left(\frac{z}{x} + 1\right) = \frac{x}{y}+\frac{y}{z+x}+\frac{z+x}{x} \geq 3\sqrt[3]{\frac{x}{y}\cdot \frac{y}{z+x}\cdot \frac{z+x}{x}} = 3$$

with equality when $x=y$ and $z = 0$.

Answer 2

By C-S and AM-GM $$\frac{x}{y}+\frac{y}{x+z}+\frac{z}{x}=\frac{x^2}{xy}+\frac{y^2}{xy+yz}+\frac{z^2}{xz}\geq\frac{(x+y+z)^2}{xy+xy+yz+xz}=$$ $$=\frac{x^2+y^2+z^2+2xy+2xz+2yz}{2xy+xz+yz}\geq \frac{4xy+2xz+2yz}{2xy+xz+yz}=2.$$ Also, we can end your work.

Indeed, we need to prove that: $$yz^2+(x^2-xy)z+x(x-y)^2\geq0,$$ which is true for $x\geq y$.

But for $x<y$ it's enough to prove that: $$x^2(x-y)^2-4xy(x-y)^2\leq0$$ or $$x(x-y)^2(x-4y)\leq0,$$ which is obvious.

Answer 3

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.

Answer 4

I found $$x^3 - 2 x^2 y + x^2 z + x y^2 - x y z + y z^2 = xy(x-y)^2+x^2(z+x-y)^2+yz^2(y+z) \geqslant 0$$