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

$2abc\cdot \dfrac{a^2}{b+c} = 2a^3\cdot \dfrac{bc}{b+c} \leq 2a^3\cdot \dfrac{b+c}{4} = \dfrac{a^3(b+c)}{2}$. 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 $\forall x,y \geq 0$