# Marian Cucoanes' Inequality With Roots and Powers

### Proof 1

Denote $a^2=x,\,$ $b^2=y,\,$ $c^2=z.\,$ By Heron's formula in expanded form,

$4S=\sqrt{(2(xy+yz+zx)-(x^2+y^2+z^2)}.$

The required inequality is equivalent to

$\displaystyle \small{(x+y+z)^2\ge[(2(xy+yz+zx)-(x^2+y^2+z^2)]\left[\frac{(x+y+z)(xy+yz+zx)}{2xyz}-\frac{3}{2}\right]}.$

WLOG, $x+y+z=3,\,$ $xy+yz+zx=3(1-t^2),\,$ $0\le t\lt 1.\,$ And, since

$0\lt (2(xy+yz+zx)-(x^2+y^2+z^2)=3(1-4t^2),$

we have $\displaystyle t\lt\frac{1}{2}.\,$ Also, due to a result by Vo Quoc Ba Can, $xyz\ge (1+t^2)(1-2t).\,$ Thus, suffice it to show that

$\displaystyle 1\ge \frac{3(1-4t^2)(1-t^2)}{2(1+t)^2(1-2t)}-\frac{1-4t^2}{2},$

which is equivalent to

$\displaystyle 3-4t^2\ge \frac{3(1+2t)(1-t)}{(1+t)},$

i.e.,

$\displaystyle 2t^2(1-2t)\ge 0,$

which is true because $\displaystyle t\lt\frac{1}{2}.$

### Proof 2

We'll use Heron's formula in the expanded form:

$16S^2=2(a^2b^2+b^2c^2+c^2a^2)-(a^4+b^4+c^4)$

which implies

(1)

$\displaystyle\left(\frac{a^2+b^2+c^2}{4S}\right)^2-3=\frac{(a^2-b^2)^2+(b^2-c^2)^2+(c^2-a^2)^2}{8S^2}.$

Further,

(2)

$\displaystyle\small{(a^2+b^2+c^2)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)^2-9=\frac{(a^2-b^2)^2}{a^2b^2}+\frac{(b^2-c^2)^2}{b^2c^2}+\frac{(c^2-a^2)^2}{c^2-a^2}}.$

The required inequality is equivalent to

$\displaystyle\left(\frac{a^2+b^2+c^2}{4S}\right)^2-3\ge\frac{1}{2}\left(a^2+b^2+c^2)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)^2-9\right)$

which (using (1),(2)) is equivalent to

$\displaystyle\frac{1}{8S^2}\sum_{cycl}(a^2-b^2)\ge\frac{1}{2}\sum_{cycl}\frac{(a^2-b^2)^2}{a^2b^2}.$

This, in turn, is equivalent to

$\displaystyle\sum_{cycl}\frac{(a^2-b^2)}{8S^2a^2b^2}(a^2b^2-4S^2)\ge 0$

which is true because $4S^2=a^2b^2\sin^2C\le a^2b^2.$

### Acknowledgment

Leo Giugiuc has kindly posted at the CutTheKnotMath facebook page the above problem, due to Marian Cucoanes, with his solution (Proof 1) and later added the solution by the problem's author (Proof 2).