Symmetric Olympiad inequality

Question

Prove that for all positive reals $a, b, c$ this inequality holds: $$\sum_{cyc}\frac{a^3}{b^2 - bc + c^2} \ge \sum_{cyc}a$$

I have proved this in a very ugly way:

I multiplied with $(a^2 - ab + b^2)(b^2 - bc + c^2)(c^2 - ca + a^2)$ and with some work i managed to prove this with Schur. I find this way boring and time-consuming. My question is - Is there a nicer solution to this?

Source: Mildorf Inequalities

Answer 1

Note by CS (with $\sum$ denoting cyclic sums): $$ \sum \frac{a^3}{b^2-bc+c^2} \geqslant \frac{\left(\sum a^2 \right)^2}{\sum a(b^2-bc+c^2)}$$

Hence it is enough to show $$\left(\sum a^2 \right)^2 \geqslant \left(\sum a\right) \cdot \sum a(b^2-bc+c^2)$$ $$\iff \sum a^4 + 2\sum a^2b^2 \geqslant \sum (2a^2b^2-a^2bc + b^3a+ ac^3)$$ which is just the fourth degree Schur inequality: $$\sum a^4 + abc\sum a \geqslant \sum ab(a^2+b^2) $$ Equality is when $a=b=c$, or when any one variable is $0$ and the other two equal.

Answer 2

Using only algebraic manipulation and the fact that a product of similarly-signed numbers is positive: \begin{gather*} \sum_\text{cyc} \frac{a^3}{b^2 - bc + c^2} - \sum_\text{cyc} a = \sum_\text{cyc} \frac{a^3(b + c) - a(b^3 + c^3)}{b^3 + c^3} \\ = \sum_\text{cyc} \frac{ab(a^2 - b^2) - ca(c^2 - a^2)}{b^3 + c^3} = \sum_\text{cyc} bc(b^2 - c^2)\left( \frac{1}{c^3 + a^3} - \frac{1}{a^3 + b^3}\right) \geqslant 0, \end{gather*} because all three terms in the final cyclic sum are non-negative.