Adil Abdullayev's Inequality With Roots and Powers


Adil Abdullayev's Inequality With Roots and Powers


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.


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.


|Contact| |Front page| |Contents| |Algebra| |Store|

Copyright © 1996-2017 Alexander Bogomolny


Search by google: