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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 $a=b=c$.

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Also posted to MO (but probably closed by now - if not, you might want to go delete it from MO). – Gerry Myerson Aug 20 '12 at 13:18
Yes, I deleted from MO – Iuli Aug 20 '12 at 13:19
up vote 3 down vote accepted

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

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


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


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

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.

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P.S. If the inequality was symmetric, it would had been a pretty standard Chebashev Sum Inequality problem, but I don't think it can be applied in this case. – N. S. Aug 20 '12 at 14:33
Last inequality does not hold: $x=1,y=2,z=3$. – sdcvvc Aug 20 '12 at 15:51
I removed that idea. – N. S. Aug 20 '12 at 23:17
The last mentioned inequality is equivalent to $\sum\frac{x}{x+2y}\ge 1$, which is true because $\sum\frac{x^2}{x^2+2xy}\ge \frac{(x+y+z)^2}{\sum x^2+2\sum xy}=1$. – Apple Nov 20 '12 at 0:24

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