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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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    @ 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:http://www.youtube.com/watch?v=v5ErHe-ls1A isn't the sticks in the video of arbitary lenghts?2012-10-12

3 Answers 3

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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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    @SouvikDey I see this is a very old post so I'm not sure if it still interests you. The inequality you mention after the substitution $2(z/x + y/z + x/y)\ge 3 + y/x + z/y + x/z$ can be proved elementarily [**see here**](http://math.stackexchange.com/a/1097350/129017) with a slight modified argument that if 2c \ngeq a \implies a > 2c \ge 2b (since, $c$ is the middle term) \displaystyle \implies \frac{1}{bc} > \frac{2}{ac} \implies \frac{4}{ab}+\frac{1}{bc} \ge \frac{2}{ac}.2015-01-09
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$\sum_{cyc}\frac{b+c-2a}{c+a}=\sum_{cyc}\frac{c-a-(a-b)}{c+a}=\sum_{cyc}(a-b)\left(\frac{1}{a+b}-\frac{1}{a+c}\right)=$ $=\sum_{cyc}\frac{(a-b)(c-b)}{(a+b)(a+c)}=\sum_{cyc}\frac{(a-b)(c-a+a-b)}{(a+b)(a+c)}=$ $=\sum_{cyc}\frac{(a-b)^2}{(a+b)(a+c)}-\sum_{cyc}\frac{(a-b)(a-c)}{(a+b)(a+c)}=$ $=\sum_{cyc}\frac{(a-b)^2}{(a+b)(a+c)}-\sum_{cyc}\frac{(a-b)(a-c)(b+c)}{\prod\limits_{cyc}(a+b)}=$ $=\sum_{cyc}\frac{(a-b)^2}{(a+b)(a+c)}-\sum_{cyc}\frac{a^2b+a^2c-2abc}{\prod\limits_{cyc}(a+b)}=$ $=\sum_{cyc}\frac{(a-b)^2(b+c)}{\prod\limits_{cyc}(a+b)}-\sum_{cyc}\frac{c(a-b)^2}{\prod\limits_{cyc}(a+b)}=\sum_{cyc}\frac{(a-b)^2b}{\prod\limits_{cyc}(a+b)}\geq0.$ Done!