Using Rearrangement Inequality . 
Let $a,b,c\in\mathbf R^+$, such that $a+b+c=3$. Prove that $$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a^2+b^2+c^2}{2}$$
$Hint$ : Use Rearrangement Inequality

My Work :-$\\$
Without Loss of Generality let's assume that $0\le a\le b\le c$ then we can infer that $c^2\ge b^2\ge a^2$ also $b+c\ge c+a \ge a+b$. Thus $\frac{1}{b+c}\le \frac{1}{c+a}\le \frac{1}{a+b}$. Hence by Rearrangement Inequality $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}$ is the greatest permutation however I am not able to express $\frac{a^2+b^2+c^2}{2}$ as another permutation of the same, hence I require assistance.
Thank You

Since triples $(a^2,b^2,c^2)$ and $\left(\frac{1}{b+c},\frac{1}{a+c},\frac{1}{a+b}\right)$ have the same ordering,
by Rearrangement and C-S we obtain: $$\sum_{cyc}\frac{a^2}{b+c}\geq\frac{1}{3}\sum_{cyc}a^2\sum_{cyc}\frac{1}{b+c}\geq\frac{1}{3}\sum_{cyc}a^2\cdot\frac{(1+1+1)^2}{\sum\limits_{cyc}(b+c)}=\frac{a^2+b^2+c^2}{2}.$$
