# $\frac{a+b-2c}{b+c} + \frac{b+c-2a}{c+a} + \frac{c+a-2b}{a+b} \geq 0$

I recently met the inequality $\frac{a+b-2c}{b+c} + \frac{b+c-2a}{c+a} + \frac{c+a-2b}{a+b} \geq 0$ , where a , b , c are all positive real numbers. I wanted to prove it but had some difficult time , seeing no connection to any known standard inequalities I began to simplify the expression multiplied by $(a+b)(b+c)(c+a)$ , after some simple and elementary but tedious calculations I obtained:-

$$(a+b)(b+c)(c+a) \left(\frac{a+b-2c}{b+c} + \frac{b+c-2a}{c+a} + \frac{c+a-2b}{a+b}\right) \\=a(a-c)^2 + b(b-a)^2 + c(c-b)^2$$ , which is obviously non-negative thus proving the inequality. But I am wondering , is there any other way(s) to prove the inequality? Also, what is the shortest way of deriving the above said identity ?

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you can think of $a,b$ and $c$ as lengths a of triangle, may be you will work it out. –  Mohamez Oct 12 '12 at 10:56
@ mohamez:- Um, a,b,c are arbitary positive real numbers , so perhaps they can not loosely be assumed to be the sides of a triangle (e.g.:- a,b,c can take the values 2,3,7 ; but a triangle can not) –  Souvik Dey Oct 12 '12 at 12:39
@ Souvik Why not?! since the sides of a triangle are also arbitary positive real numbers. for example take $3$ sticks of lenghts respectively $2m$, $3m$ and $7m$ can you join the vertices to make up a triangle? i'm making no condition on the triangle! Whatch this:youtube.com/watch?v=v5ErHe-ls1A isn't the sticks in the video of arbitary lenghts? –  Mohamez Oct 12 '12 at 18:17

We can set $a+b=C$ and so on, then minimize

$$f(A,B,C)=\sum_{cyc}\frac{2C-B}{A}.$$

If we regard $f$ as a function of $B$ and $C$ only, we have that $\frac{\partial f}{\partial B}=0$ implies $(B^2-AC)(2A-C)=0$, and $\frac{\partial f}{\partial C}=0$ implies $(C^2-AB)(2B-A)=0$. So we have four stationary points:

$$(B/A,C/A)\in\left\{(1,1),(2,4),(2,1/2),(1/2,1/4)\right\}.$$

Without loss of generality we can additionally assume $A\leq B\leq C$, having stationary points for

$$(A,B,C) = (\lambda,\lambda,\lambda)\quad\mbox{or}\quad(\lambda,2\lambda,4\lambda).$$

We can now substitute these values into $f$ to prove the inequality.

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To address the first question, let $x = b+c$, $y = c+a$ and $z = a+b$. Then $2c=x+y-z$, so the inequality rewrites as $$\frac{y-(x+y-z)}{x} + \frac{z-(y+z-x)}{y} + \frac{x-(x+y-z)}{z} \ge 0$$ or $$\frac{z}{x} + \frac{x}{y} + \frac{y}{z} \ge 3,$$ which follows immediately from the AM-GM inequality.

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You have mistaken in writing down the inequality , what you have wrote is ((a-c)/(b+c)) + ((b-a)/(c+a)) + ((c-b)/(a+b)) ≥ 0 ; this is not what I stated. Using x=b+c , y=c+a , z=a+b , the inequality I posed becomes 2(z/x + y/z + x/y)≥ 3 + y/x + z/y + x/z ; which is much stronger than AM-GM inequality –  Souvik Dey Oct 17 '12 at 9:35
I totally agree with @Souvik Dey - since the inequality has more than 1 stationary points, it cannot follow from an inequality among means, since these inequalities have the unique stationary point $a=b=\ldots$. –  Jack D'Aurizio Oct 23 '12 at 10:28
@JackD'Aurizio:- Nice to see that you noticed the fact that v_Enhance's proof is wrong ; though I have accepted your previous answer I still think that the procedure you adopted is not quite elementary as the inequality itself. Can you give it a try of proving the inequality by proving the alternative version that I stated in my last comment , in an elementary way ? –  Souvik Dey Oct 23 '12 at 12:47