The least positive real number $k$ for which $k\left(\sqrt{(x+y)(y+z)(z+x)}\right)\geq x\sqrt{y}+y\sqrt{z}+z\sqrt{x}$

Question

The least positive real number $k$ for which $$k\left(\sqrt{(x+y)(y+z)(z+x)}\right)\geq x\sqrt{y}+y\sqrt{z}+z\sqrt{x}$$

Where $x,y,z>0$

$\bf{My\; Try::}$ Here $$k\geq \frac{x\sqrt{y}+y\sqrt{z}+z\sqrt{x}}{\sqrt{(x+y)(y+z)(z+x)}}$$

Using $\bf{A.M\geq G.M}$ Inequality

$$(x+y)(y+z)(z+x)\geq 8xyz$$

How can i solve it after that, Help required, Thanks

Answer 1

Let $x=y=z$.

Hence, $k\geq\frac{3}{2\sqrt2}$.

We'll prove that $k=\frac{3}{2\sqrt2}$ is valid.

Indeed, we need to prove that

$9(x+y)(y+z)(z+x)\geq8(x\sqrt{y}+y\sqrt{z}+z\sqrt{x})^2$.

By C-S $(x+y+z)(xy+yz+zx)\geq(x\sqrt{y}+y\sqrt{z}+z\sqrt{x})^2$.

Thus, it remains to prove that $$9(x+y)(y+z)(z+x)\geq8(x+y+z)(xy+yz+zx)$$ or $$\sum_{cyc}z(x-y)^2\geq0$$ Done!