2
$\begingroup$

Let $a,b,c$ be the side lengths of a triangle. Prove that

$\frac{a}{3a-b+c}+\frac{b}{3b-c+a}+\frac{c}{3c-a+b} \geq 1 . $

I found this inequlity in the chapter entitled Cauchy-Schwarz, but I cannot find a proof for this inequality. I used the triangle inequality and Cauchy-Schwarz but I proved the case of equality; that is for $a=b=c$.

  • 0
    Yes, I deleted from MO2012-08-20

1 Answers 1

3

Here is an attempt.

It is well known that $a,b,c$ are the sides of a triangle if and only if you can find numbers $x,y,z >0$ so that $a=x+y, b=x+z, c=y+z$. Your inequality becomes then

$\frac{x+y}{2x+4y}+\frac{x+z}{4x+2z}+\frac{y+z}{4z+2y} \geq 1 \,;\, \forall x,y,z >0 \,.$

This inequality reduces after horrible computations to

$ x^2y+y^2z+z^2x \geq 3xyz $

But this is a bad solution.

Here is a better idea, cannot complete the solution though:

The equation is equivalent to

$\sum_{cyc}\frac{x+y}{x+2y} \geq 2$

or

$\sum_{cyc}1-\frac{y}{x+2y} \geq 2$

or

$1 \geq \sum_{cyc}\frac{y}{x+2y} \,.$

Probably the easiest approach from here would be to denote $x+2y=m, y+2z=n, z+2x=p$ and solve for $x,y,z$. This suggest that probably it would had been best to denote $m=3a-b+c, n=3b-c+a, p=3c-a+b$ from beginning.

  • 2
    The last mentioned inequality is equivalent to $\sum\frac{x}{x+2y}\ge 1$, which is true because $\sum\frac{x^2}{x^2+2xy}\ge \frac{(x+y+z)^2}{\sum x^2+2\sum xy}=1$.2012-11-20