parmenides51 wrote:
Let $x,y,z$ be a positive real numbers. Prove that $$\frac {x}{\sqrt {x + y}} + \frac {y}{\sqrt {y + z}} + \frac { z}{\sqrt {z + x}}\geq\sqrt [4]{\frac {27(yz + zx + xy)}{4}}$$
(dektep)
We have
$$\frac {x}{\sqrt {x + y}} + \frac {y}{\sqrt {y + z}} + \frac { z}{\sqrt {z + x}}
Proof 1: Note that $u \mapsto \frac{1}{\sqrt u}$ is convex on $u > 0$.
Proof 2: The following inequality is known: Let $a, b, c, x, y, z > 0$ and $\lambda > 0$. Then
$$\frac{x^{\lambda+1}}{a^\lambda} + \frac{y^{\lambda+1}}{b^\lambda} + \frac{z^{\lambda+1}}{c^\lambda}\ge \frac{(x+y+z)^{\lambda+1}}{(a+b+c)^\lambda}.$$
Proof 3: Use Holder.
It suffices to prove that
$$\frac{(x+y+z)^{3/2}}{\sqrt{x(x+y) + y(y+z) + z(z+x)}}
\ge \sqrt [4]{\frac {27(yz + zx + xy)}{4}}$$which is true (easy).
We are done.