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Let $a$, $b$ and $c$ be the three sides of a triangle. Show that $\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3$

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    $\begingroup$ First thing you should try is Ravi substitution. $\endgroup$ – punctured dusk Feb 19 '15 at 12:34
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    $\begingroup$ See also this. This question is different, but most likely it can be solved analoguously. $\endgroup$ – punctured dusk Feb 19 '15 at 12:39
  • $\begingroup$ @barto Can you please kindly help? I've tried using Ravi substitution. But I'm not able to reach the final output. $\endgroup$ – N S Feb 19 '15 at 12:39
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    $\begingroup$ The example given by @barto is useful, as using rearrangement inequality, one can show $$\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c} \ge \frac{a}{c+a-b}+\frac{b}{a+b-c}+\frac{c}{b+c-a}$$ $\endgroup$ – Macavity Feb 19 '15 at 13:12
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    $\begingroup$ Assume the denominators are positive numbers $x,y,z$ respectively. Can you express $a,b,c$ in terms of $x,y,z$? Then AM-GM will be your friend. $\endgroup$ – Macavity Nov 29 '16 at 9:29
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$a, b, c$ are sides of a triangle iff there exists positive reals $x, y, z$ s.t. $a=x+y, b=y+z, c = z+x$. In terms of these variables, the inequality is $$\sum_{cyc} \frac{a}{b+c-a} = \sum_{cyc} \frac{x+y}{2z} \ge 3$$

Now the last is easy to show with AM-GM of all $6$ terms. $$\sum_{cyc} \frac{x+y}{2z} = \frac12\left(\frac{x}z+\frac{y}z+\frac{y}x+\frac{z}x+\frac{z}y+\frac{x}y \right) \ge \frac12\left(6\sqrt[6]{\frac{x}z\cdot\frac{y}z\cdot\frac{y}x\cdot\frac{z}x\cdot\frac{z}y\cdot\frac{x}y} \right) = 3$$

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  • $\begingroup$ I'm not getting anything. How can you say that a,b,c is related to x,y,z? What's the intuition? $\endgroup$ – N S Feb 19 '15 at 12:46
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    $\begingroup$ $x = \frac{c + a - b}{2}$, $y = \frac{a + b - c}{2}$, $z = \frac{b + c - a}{2}$ $\endgroup$ – Christopher Feb 19 '15 at 12:50
  • $\begingroup$ Did you see the Ravi link? Draw an incircle - the circle that touches all three sides of the triangle. This cuts the $a,b,c$ sides into $x+y,y+z,z+x$ respectively. $\endgroup$ – Empy2 Feb 19 '15 at 12:51
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    $\begingroup$ @NS We know by triangle inequality, $a+b-c, b+c-a, c+a-b$ must be non-negative / positive if you have non-degenerate triangles. $x, y, z$ are simply non-negative / positive variables which capture the triangle inequality. $\endgroup$ – Macavity Feb 19 '15 at 12:54
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    $\begingroup$ @NS added that in the post... $\endgroup$ – Macavity Feb 19 '15 at 13:04
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Set $b+c-a=x, a+c-b=y$ and $a+b-c=z$
Due to triangular inequality, $x,y,z$ are positive. Hence we can apply the AM-GM inequality. We have $x+y=b+c-a+a+c-b=2c \implies c = \frac{x+y}{2}$
Similarly $a = \frac{y+z}{2}$ and $b = \frac{z+x}{2}$
Now we have, $$\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}=\frac{y+z}{2x}+\frac{z+x}{2y}+\frac{x+y}{2z}$$ $$\Leftrightarrow 2\left ( \frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c} \right )=\left ( \frac{x}{y}+\frac{y}{x} \right )+\left ( \frac{y}{z}+\frac{z}{y} \right )+\left ( \frac{z}{x}+\frac{x}{z} \right )\geq 6$$ Hence $\ $ $\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\geq 3$

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  • $\begingroup$ so this has nothing to do with triangles? $\endgroup$ – MichaelChirico Nov 29 '16 at 13:48
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    $\begingroup$ Of course it has to do something with triangles. The positivity of x,y,z are due to triangular inequality. $\endgroup$ – Shraddheya Shendre Nov 29 '16 at 14:06
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By C-S $\sum\limits_{cyc}\frac{a}{b+c-a}=\sum\limits_{cyc}\frac{a^2}{ab+ac-a^2}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(2ab-a^2)}\geq3$, where the last inequality it's just $$\sum\limits_{cyc}(a-b)^2\geq0.$$ Done!

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Suppose that $S>0$. Then for $x\in(0,S/2)$, the function $f(x)=\frac{x}{S-2x}$ is convex. Thus, by Jensen's inequality and with $S=a+b+c$, we have $$ \frac{1}{3}f(a)+\frac{1}{3}f(b)+\frac{1}{3}f(c)\geq f\left(\frac{1}{3}(a+b+c)\right)=\frac{S/3}{S-2S/3}=1. $$ This is equivalent to $f(a)+f(b)+f(c)\geq 3$, which is your inequality.

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$\sum\limits_{cyc}\frac{a}{b+c-a}-3=\sum\limits_{cyc}\left(\frac{a}{b+c-a}-1\right)=\sum\limits_{cyc}\frac{a-b-(c-a)}{b+c-a}=$ $=\sum\limits_{cyc}(a-b)\left(\frac{1}{b+c-a}-\frac{1}{a+c-b}\right)=\sum\limits_{cyc}\frac{2(a-b)^2}{(b+c-a)(a+c-b)}\geq0$

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Since $(a,b,c)$ and $\left(\frac{1}{b+c-a},\frac{1}{a+c-b},\frac{1}{a+b-c}\right)$ are the same ordered, by Chebyshov and C-S we obtain: $\sum\limits_{cyc}\frac{a}{b+c-a}\geq\frac{1}{3}(a+b+c)\sum\limits_{cyc}\frac{1}{b+c-a}=\frac{1}{3}\sum\limits_{cyc}(b+c-a)\sum\limits_{cyc}\frac{1}{b+c-a}\geq\frac{1}{3}\cdot9=3$.

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    $\begingroup$ Someone's pushing for karma points here w/ 3 separate answers :-) . $\endgroup$ – Carl Witthoft Nov 29 '16 at 13:09
  • $\begingroup$ Dear @Carl Witthoft You are welcome to show us your proof ;) $\endgroup$ – Michael Rozenberg Nov 29 '16 at 13:39
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Since $a,b,c$ are sides of a triangle, we can set $a = x+y$, $b = x+z$, $c = y+z$.

Plugging that in gives

\begin{align} \frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c} &= \frac{x+y}{2z} + \frac{x+z}{2y} + \frac{y+z}{2x} \\ &= \frac{xy(x+y) + xz(x+z) + yz(y+z)}{2xyz}\\ &= \frac{x^2y+ xy^2 + x^2z + xz^2 + y^2z + yz^2}{2xyz}\\ &\ge \frac{3}{xyz}\sqrt[6]{x^6y^6z^6}\\ &= 3 \end{align}

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