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Let $a$, $b$ and $c$ be a sides-lengths of the triangle. Prove that: $$\frac{a}{(a+b)^2}+\frac{b}{(b+c)^2}+\frac{c}{(c+a)^2}\geq\frac{9}{4(a+b+c)}$$

The Buffalo way kills it, but I am looking for a nice proof for this nice inequality.

SOS (sums of squares) gives $\sum\limits_{cyc}\frac{(a-b)^2(ab+b^2-ac)}{(a+b)^2}\geq0$ and I don't see what comes next.

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    Just out of curiosity. What is BW and SOS? – 2017-01-23
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    I would try the following. Assume $a+b+c=1$ due to homogeneity. Try to use rearrangement inequality to transform your inequality into $a/(1-a)^2+b/(1-b)^2+c/(1-c)^2 \geq 9/4$. Now, this is (almost) obvious. I am not sure the second step works. – 2017-01-23
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    @ivan about BW and SOS see here: http://www.artofproblemsolving.com/community/c6h522084 and here: http://www.artofproblemsolving.com/community/c6h80127 – 2017-01-24
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    @Takahiro Waki I think he got another inequality. – 2017-01-27

3 Answers 3

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clearing the denominators you inequality is equivalent to $$4\,b{a}^{5}+7\,{b}^{2}{a}^{4}+2\,bc{a}^{4}-5\,{c}^{2}{a}^{4}-2\,{b}^{3 }{a}^{3}+2\,{b}^{2}c{a}^{3}-6\,b{c}^{2}{a}^{3}-2\,{c}^{3}{a}^{3}-5\,{b }^{4}{a}^{2}-6\,{b}^{3}c{a}^{2}-6\,{b}^{2}{c}^{2}{a}^{2}+2\,b{c}^{3}{a }^{2}+7\,{c}^{4}{a}^{2}+2\,{b}^{4}ca+2\,{b}^{3}{c}^{2}a-6\,{b}^{2}{c}^ {3}a+2\,b{c}^{4}a+4\,{c}^{5}a+4\,{b}^{5}c+7\,{b}^{4}{c}^{2}-2\,{b}^{3} {c}^{3}-5\,{b}^{2}{c}^{4} \geq 0$$ now we use the Ravi-substitution $$a=z+y,b=x+z,c=x+y$$ and your inequality is equivalent to $$4\,{x}^{6}-4\,{x}^{5}y+36\,{x}^{5}z-21\,{x}^{4}{y}^{2}+22\,{x}^{4}yz+ 67\,{x}^{4}{z}^{2}+22\,{x}^{3}{y}^{3}-62\,{x}^{3}{y}^{2}z-14\,{x}^{3}y {z}^{2}+22\,{x}^{3}{z}^{3}+67\,{x}^{2}{y}^{4}-14\,{x}^{2}{y}^{3}z-150 \,{x}^{2}{y}^{2}{z}^{2}-62\,{x}^{2}y{z}^{3}-21\,{x}^{2}{z}^{4}+36\,x{y }^{5}+22\,x{y}^{4}z-62\,x{y}^{3}{z}^{2}-14\,x{y}^{2}{z}^{3}+22\,xy{z}^ {4}-4\,x{z}^{5}+4\,{y}^{6}-4\,{y}^{5}z-21\,{y}^{4}{z}^{2}+22\,{y}^{3}{ z}^{3}+67\,{y}^{2}{z}^{4}+36\,y{z}^{5}+4\,{z}^{6} \geq 0$$ now we use $$y=x+u,z=x+u+v$$ and we get the equivalent inequality $$\left( 512\,{u}^{2}+512\,uv+512\,{v}^{2} \right) {x}^{4}+ \left( 1408 \,{u}^{3}+2688\,{u}^{2}v+2560\,u{v}^{2}+640\,{v}^{3} \right) {x}^{3}+ \left( 1440\,{u}^{4}+4032\,{u}^{3}v+4704\,{u}^{2}{v}^{2}+2112\,u{v}^{ 3}+288\,{v}^{4} \right) {x}^{2}+ \left( 648\,{u}^{5}+2384\,{u}^{4}v+ 3456\,{u}^{3}{v}^{2}+2224\,{u}^{2}{v}^{3}+616\,u{v}^{4}+56\,{v}^{5} \right) x+108\,{u}^{6}+492\,{u}^{5}v+867\,{u}^{4}{v}^{2}+730\,{u}^{3} {v}^{3}+307\,{u}^{2}{v}^{4}+60\,u{v}^{5}+4\,{v}^{6} \geq 0$$ which is true.

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    Sonnhard, I proved this inequality by BW. See my words above. – 2017-04-08
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    I didn't read this sorry Michael – 2017-04-08
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Have: $\Leftrightarrow \left(\sum _{cyc}a\right)\left(\sum _{cyc}\frac{a}{\left(b+c\right)^2}\right)\ge \frac{9}{4}$

By C-S and Nesbitt : $LHS\ge \left(\frac{a}{b+c}+\frac{b}{b+c}+\frac{c}{c+a}\right)^2=\frac{9}{4}$

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    See please better the given You proved another inequality. – 2017-07-05
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WLOG $a+b+c=1$

$f\left(0;1\right)\rightarrow \left(0;\infty \right);f\left(x\right)=\frac{x}{\left(1-x\right)^2}$

See $f$ is a convex function, then use Jensen's inequality

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    You still think about another inequality. See better the given. – 2017-07-06