Prove or disprove inequality $\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\le\frac{a^4+b^4+c^4}{2abc}$.

Question

Let $a$, $b$ and $c$ be real numbers greater than $0$. Prove inequality $$\displaystyle{\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\le\frac{a^4+b^4+c^4}{2abc}}.$$

Answer 1

$2abc\cdot \dfrac{a^2}{b+c} = 2a^3\cdot \dfrac{bc}{b+c} \leq 2a^3\cdot \dfrac{b+c}{4} = \dfrac{a^3(b+c)}{2}$. Thus:

$2abc\cdot \dfrac{a^2}{b+c} + 2abc\cdot \dfrac{b^2}{c+a} + 2abc\cdot \dfrac{c^2}{a+b} \leq \dfrac{(a^3b + ab^3) + (a^3c + ac^3) + (b^3c + bc^3)}{2} \leq \dfrac{(a^4 + b^4) + (b^4 + c^4) + (c^4 + a^4)}{2} = a^4 + b^4 + c^4$, and this is true because:

$x^3y + xy^3 \leq x^4 + y^4 \iff (x^3 - y^3)(x - y) \geq 0$ is true $\forall x,y \geq 0$

Answer 2

By AM-GM and C-S we obtain: $$\frac{a^4+b^4+c^4}{2abc}=\sum_{cyc}\frac{a^3}{2bc}\geq\sum_{cyc}\frac{2a^3}{(b+c)^2}=$$ $$=\frac{2}{a+b+c}\sum_{cyc}\frac{a^3}{(b+c)^2}\sum_{cyc}a\geq\frac{2}{a+b+c}\left(\sum_{cyc}\frac{a^2}{b+c}\right)^2=$$ $$=\frac{2}{a+b+c}\cdot\sum_{cyc}\frac{a^2}{b+c}\cdot\sum_{cyc}\frac{a^2}{b+c}\geq$$ $$\geq\frac{2}{a+b+c}\cdot\frac{(a+b+c)^2}{2(a+b+c)}\cdot\sum_{cyc}\frac{a^2}{b+c}=\sum_{cyc}\frac{a^2}{b+c}.$$ Done!