Inequality $\sqrt{xy+yz+zx} \ge \frac {8}{15} (x+y+z)$

Question

By Titu's inequality:

$\sum_{cyc} \frac {1}{x+y} \ge \frac {(1+1+1)^2}{2(x+y+z)} = \frac {9}{2(x+y+z)}$

Then, to prove:

$ \frac {3}{x+y+z} + \sum_{cyc} \frac {1}{x+y} \ge \frac {4}{\sqrt{xy+yz+zx}}$

it is enough that to prove

$\frac {15}{2(x+y+z)} \ge \frac {4}{\sqrt{xy+yz+zx}}$

or

$\sqrt{xy+yz+zx} \ge \frac {8}{15} (x+y+z)$

and i stuck. Thank you

Answer 1

WARNING.- This is not an inequality to prove but to solve (find $(x,y,z)$ such that .....).

In fact, for $(x,y,z)=(10,10,1)$ we would have $$10.9544\ge11.2$$

Answer 2

If I understood you right then we need to prove that: $$\frac{3}{x+y+z}+\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{y+z}\geq\frac{4}{\sqrt{xy+xz+yz}},$$ where $x$, $y$ and $z$ are non-negatives such that $xy+xz+yz\neq1.$

If so, it follows from C-S and AM-GM.

Indeed, since our inequality is homogeneous, we can assume that $xy+xz+yz=1$.

Thus, $$\frac{3}{x+y+z}+\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{y+z}=$$ $$=\frac{3}{x+y+z}+\sum_{cyc}\frac{xy+xz+yz}{x+y}=$$ $$=\frac{3}{x+y+z}+x+y+z+\sum_{cyc}\frac{xy}{x+y}=$$ $$=\frac{3}{x+y+z}+x+y+z+\sum_{cyc}\frac{x^2y^2}{x^2y+y^2x}\geq$$ $$\geq\frac{3}{x+y+z}+x+y+z+\frac{(xy+xz+yz)^2}{\sum\limits_{cyc}(x^2y+y^2x)}\geq$$ $$\geq\frac{3}{x+y+z}+x+y+z+\frac{(xy+xz+yz)^2}{\sum\limits_{cyc}(x^2y+y^2x+xyz)}=$$ $$=\frac{3}{x+y+z}+x+y+z+\frac{(xy+xz+yz)^2}{\sum\limits_{cyc}(xy+xz+yz)(x+y+z)}=$$ $$=\frac{4}{x+y+z}+x+y+z\geq2\sqrt{\frac{4}{x+y+z}\cdot(x+y+z)}=$$ $$=4=\frac{4}{\sqrt{xy+xz+yz}}$$ and we are done!