Adil Abdullayev's Inequality With Roots and Powers

Proof

We know that $3\sqrt{3}\ge8\sin A\sin B\sin C\,$ which is equivalent to $\displaystyle 3\sqrt{3}(abc)^2\ge\frac{(abc)^3}{R^3}\,$ which, in turn, is equivalent to

$\sqrt{3}(abc)^{2/3}\ge 4S.$

Denote $a^2=x,\,$ $b^2=y,\,$ and $c^2=z.\,$ In terms of $x,y,z,\,$ suffice it to show that

$\displaystyle x+y+z\ge\sqrt{3}(xyz)^{1/3}\sqrt[4]{(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}$

or, equivalently,

$\displaystyle (x+y+z)^3\ge 9(xyz)^{1/3}(xy+yz+zx).$

But $x+y+z\ge 3(xyz)^{1/3}\,$ and $(x+y+z)^2\ge 3(xy+yz+zx),\,$ which completes the proof.

Acknowledgment

Leo Giugiuc has kindly posted at the CutTheKnotMath facebook page the above problem, due to Adil Abdullayev, with his solution. The problem has been originally posted at the mathematical inequalities facebook group.