<s>[INST] The least positive real number $k$ for which $k\left(\sqrt{(x+y)(y+z)(z+x)}\right)\geq x\sqrt{y}+y\sqrt{z}+z\sqrt{x}$ 
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
 [/INST] 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!
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