by henderson, Sep 29, 2016, 12:23 PM 
Let 
be positive real numbers such that 
Prove that![\[\frac{a^2(b+c)}{b^2+c^2}+\frac{b^2(c+a)}{c^2+a^2}+\frac{c^2(a+b)}{a^2+b^2}\geq \sqrt{3(a+b+c)}.\]](http://latex.artofproblemsolving.com/1/5/2/1524a11c1e9503f59895a6ca7fa9c6ef9c2121fd.png)
"Do not worry too much about your difficulties in mathematics, I can assure you that mine are still greater." - Albert Einstein
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Firstly, let us write the given inequality as the following:![\[(a^2+b^2+c^2)\left(\frac{b+c}{b^2+c^2}+\frac{c+a}{c^2+a^2}+\frac{a+b}{a^2+b^2}\right)
\geq 2(a+b+c)+\sqrt{3(a+b+c)}. \quad \quad \quad \color{blue}{(1)} \]](http://latex.artofproblemsolving.com/0/c/a/0ca94591543ddebcb19368ed529e42b0d9e83f90.png)
If ![\[\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b}\leq \frac{3(a^2+b^2+c^2)}{a+b+c}\]](http://latex.artofproblemsolving.com/c/9/f/c9fe0197dd833f8ff10e9fafe7b822a101d5deec.png)
holds.
![\[\sum_{cyc}\frac{\left((a+b)^2-2ab\right)(a+b+c)}{a+b} \leq 3(a^2+b^2+c^2)\]](http://latex.artofproblemsolving.com/6/1/3/6134262c12bfe2f5a674687e1fae10ece81fde72.png)
![\[a^2+b^2+c^2 + \sum_{cyc}\frac{2abc}{a+b} \geq 2(ab+bc+ca).\]](http://latex.artofproblemsolving.com/d/2/0/d20442c4c0d1222df65b94a59624a95c72151a70.png)
The last one can be proved just using 
and ![\[a^2+b^2+c^2 + \frac{9abc}{a+b+c} \geq 2(ab+bc+ca)\]](http://latex.artofproblemsolving.com/5/1/1/511ff254f66104a8b0865cee9fed486f8401f20b.png)
for all 
![\[a^2+b^2+c^2 + \sum_{cyc}\frac{2abc}{a+b} \geq a^2+b^2+c^2+\frac{9\cdot 2abc}{2(a+b+c)}=a^2+b^2+c^2 + \frac{9abc}{a+b+c}\geq 2(ab+bc+ca),\]](http://latex.artofproblemsolving.com/d/a/b/dab1e0a1af3372fb70285b0a7659db7d602e7ced.png)
as desired. 
![\[9\leq\sum_{cyc}{\frac{b+c}{b^2+c^2}} \cdot \sum_{cyc}{\frac{b^2+c^2}{b+c}}\leq \sum_{cyc}{\frac{b+c}{b^2+c^2}} \cdot \frac{3(a^2+b^2+c^2)}{a+b+c},\]](http://latex.artofproblemsolving.com/d/4/e/d4ed802db06a67af6c6897ca0cadc34af39b6a89.png)
thus we have![\[\sum_{cyc}{\frac{b+c}{b^2+c^2}}\geq \frac{3(a+b+c)}{a^2+b^2+c^2}\]](http://latex.artofproblemsolving.com/1/c/7/1c799729a52a2ef94425ede265f4fd0db30e0432.png)
![\[(a^2+b^2+c^2)\left(\frac{b+c}{b^2+c^2}+\frac{c+a}{c^2+a^2}+\frac{a+b}{a^2+b^2}\right)\geq 3(a+b+c). \quad \quad \quad \color{blue}{(2)} \]](http://latex.artofproblemsolving.com/c/6/5/c65a1f6c85f3455e32c423407de456c058e992df.png)
Applying 
to 
it suffices to show that![\[a+b+c\geq \sqrt{3(a+b+c)}.\]](http://latex.artofproblemsolving.com/2/1/b/21b4946a27a7d59cb30473ba64efc70074db3d36.png)
But this one is straightforward:![\[a+b+c=\sqrt{a+b+c} \cdot \sqrt{a+b+c}\geq \sqrt{3\sqrt[3]{abc}} \cdot \sqrt{a+b+c}=\sqrt{3(a+b+c}.\]](http://latex.artofproblemsolving.com/7/a/b/7ab24697e239eb800269fefb9e5f90cbfc3d91fc.png)
The proof is completed.