# 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

• IMO, the first line after "My Try" is the final answer. You could try squaring both sides, but introducing more inequalities will only give some $k$, not the least positive real number $k$. – Simply Beautiful Art Nov 1 '16 at 13:26
• If you mean it to hold for all $x,y,z>0$, I am afraid there is a problem. Take $x\to \infty$, the LHS is asymptotically equivalent to $2 k \sqrt{x}$ while the RHS is equivalent to $x\sqrt{y}$... which is much larger than the LHS... so I am wondering if you are not talking about, rather, the reverse inequality. – Pablo Rotondo Nov 1 '16 at 13:28
• Sorry Pablo Rotondo I have edited my question – juantheron Nov 1 '16 at 13:32
• Looks better now! – Pablo Rotondo Nov 1 '16 at 13:35

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!

• @juantheron What do you think about my proof. Something is not right? – Michael Rozenberg Nov 2 '16 at 6:41