Prove/disprove that $\frac{x^2 - \sqrt{yz}}{yz - x} + \frac{y^2 - \sqrt{zx}}{zx - y} + \frac{z^2 - \sqrt{xy}}{xy - z} \ge 0$

Question

Prove/disprove that

$$ \frac{x^2 - \sqrt{yz}}{yz - x} + \frac{y^2 - \sqrt{zx}}{zx - y} + \frac{z^2 - \sqrt{xy}}{xy - z} \ge 0$$ with $x$, $y$ and $z$ are positives.

I tried to use the Schur's inequality for this but it didn't help.

If the inequality is not correct, what minor change could have been done to the inequality so that it is correct for $\forall x, y, z \in \mathbb R^+|yz - x, zx - y, xy - z \ne 0$?

Answer 1

Prove/disprove that $$\dfrac{x^2 - \sqrt{yz}}{yz - x} + \dfrac{y^2 - \sqrt{zx}}{zx - y} + \dfrac{z^2 - \sqrt{xy}}{xy - z} \ge 0$$with $x$, $y$ and $z$ are positives.

Choosing for example $(x,y,z)=(1,1,4)$ we get $$\frac{1^2-\sqrt{4}}{4-1}+\frac{1^2-\sqrt{4}}{4-1}+\frac{4^2-\sqrt{1}}{1-4}=-\frac{1}{3}-\frac{1}{3}-\frac{15}{3}=-\frac{17}{3}<0,$$ so the inequality does not hold.

If the inequality is not correct, what minor change could have been done to the inequality so that it is correct for $\forall x, y, z \in \mathbb Z^+|yz - x, zx - y, xy - z \ne 0$?

Sure, we can change it for example to

$$\dfrac{x^2 - \sqrt{yz}}{yz + x} + \dfrac{y^2 - \sqrt{zx}}{zx + y} + \dfrac{z^2 - \sqrt{xy}}{xy + z} \ge 0.$$ (Notice that condition $yz - x, zx - y, xy - z \ne 0$ can be now dropped). This holds since for example for $x\geq1$ we have $x^2\geq x$, also without loss of generality let's $x \leq y \leq z$, and so we have $$ \sum \dfrac{x^2 - \sqrt{yz}}{yz + x} \geq \frac{1}{z^2+z} \sum (x - \sqrt{yz}) $$ and sum on the right is non-negative due to $x+y+z\geq \sqrt{xy}+\sqrt{yz}+\sqrt{zx}$, which is just rearrangement inequality applied on $\sqrt{x} \leq \sqrt{y} \leq \sqrt{z}$.

Answer 2

It is not true, try $$x=1,y=2,z=3$$, we get $$-{\frac{24}{5}}-1/5\,\sqrt {2}\sqrt {3}-\sqrt {3}+\sqrt {2}$$, this is negative.