### Problem

Prove that in any $\Delta ABC,\,$ with side lengths $a,b,c\,$

\displaystyle\begin{align} &(a+b-c)(b+c-a)(c+a-b)\\ &\qquad\qquad\le 2\min\{a\cdot\frac{b^2c^2}{b^2+c^2},b\cdot\frac{c^2a^2}{c^2+a^2},c\cdot\frac{a^2b^2}{a^2+b^2}\}\\ &\qquad\qquad\le 2\max\{a\cdot\frac{b^2c^2}{b^2+c^2},b\cdot\frac{c^2a^2}{c^2+a^2},c\cdot\frac{a^2b^2}{a^2+b^2}\}\\ &\qquad\qquad\le abc. \end{align}

### Solution

We first note that

(1)

$\displaystyle (a+b-c)(b+c-a)(c+a-b)\le 2a\cdot\frac{b^2c^2}{b^2+c^2}.$

Indeed, with Ravi's substitution, $a=y+z,\,$ $b=z+x,\,$ $c=x+y,\,$ (1) reduces to

$\displaystyle 8xyz\le 2(y+z)\frac{(z+x)^2(x+y)^2}{(z+x)^2+(x+y)^2},$

or, equivalently, to

$\displaystyle 4xyz\cdot\frac{(z+x)^2+(x+y)^2}{(z+x)^2(x+y)^2}\le y+z.$

But

\displaystyle\begin{align} xyz\cdot\frac{(z+x)^2+(x+y)^2}{(z+x)^2(x+y)^2}&=4xyz\cdot\left(\frac{1}{(x+y)^2}+\frac{1}{(z+x)^2}\right)\\ &\le 4xyz\cdot\left(\frac{1}{4xy}+\frac{1}{4zx}\right)=y+z, \end{align}

with equality for $x=y=z,\,$ i.e., $a=b=c.$

Due to the symmetry of the left-hand side in (1),

$\displaystyle (a+b-c)(b+c-a)(c+a-b)\le 2\min_{cycl}\{a\cdot\frac{b^2c^2}{b^2+c^2}\}.$

For the second inequality, observe that $\displaystyle 2a\frac{b^2c^2}{b^2+c^2}\le 2a\frac{b^2c^2}{2bc}=abc.$ Due to the symmetry of the right-hand side,

$\displaystyle 2\max_{cycl}\{a\cdot\frac{b^2c^2}{b^2+c^2}\}\le abc.$

### Acknowledgment

The statement is due to Dorin Marghidanu as is the above solution. Marian Dinca has kindly alerted me to the original post and supplied a solution of his own that was a little more complicated than the above.

Obviously, the statement is a refinement of Padoa's inequality.

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