Show $ \frac {a^3}{a^2+b^2+c^2-bc}+\frac {b^3}{a^2+b^2+c^2-ca} +\frac {c^3}{a^2+b^2+c^2-ab} \le \frac {(a+b+c)^2 (a^4+b^4+c^4)}{6abc(ab+bc+ca)} $

Question

If $a,b,c$ are positive real numbers then how do we prove that $ \dfrac {a^3}{a^2+b^2+c^2-bc}+\dfrac {b^3}{a^2+b^2+c^2-ca} +\dfrac {c^3}{a^2+b^2+c^2-ab} \le \dfrac {(a+b+c)^2 (a^4+b^4+c^4)}{6abc(ab+bc+ca)} $

Please help. Thanks

Answer 1

Noting $ab+bc+ca \le a^2+b^2+c^2$ and $a^4+b^4+c^4 \ge \frac13(a^2+b^2+c^2)^2$, it is sufficient to show:

$$\sum_{cyc} \frac{a^3}{a^2+b^2+c^2-bc} \le \frac{(a+b+c)^2(a^2+b^2+c^2)}{18 abc}$$

Now $a^2+(b^2+c^2-bc) \ge a^2+bc$. So it is enough to show $$ \frac{(a+b+c)^2(a^2+b^2+c^2)}{18 abc} \ge \sum_{cyc} \frac{a^3}{a^2+bc} = (a+b+c)-abc\sum_{cyc}\frac1{a^2+bc}$$ where we have used $\dfrac{a^3}{a^2+bc} = a - \dfrac{abc}{a^2+bc}$.

This can be written as a quadratic in $\frac{a+b+c}{abc}$: $$ (a^2+b^2+c^2)\left(\frac{a+b+c}{abc}\right)^2-18\left(\frac{a+b+c}{abc}\right)+18\sum_{cyc}\frac1{a^2+bc} \ge 0$$

and the discriminant condition is then: $$(a^2+b^2+c^2)\sum_{cyc}\frac1{a^2+bc} \ge \tfrac92$$

which follows readily from Cauchy-Schwarz inequality.

Answer 2

$(a+b+c)^2\geq3(ab+ac+bc)$.

Thus, it remains to prove that $$\sum_{cyc}\frac{a^3}{a^2+b^2+c^2-bc}\leq\frac{a^4+b^4+c^4}{2abc}$$ or $$\sum_{cyc}\left(\frac{a^3}{2bc}-\frac{a^3}{a^2+b^2+c^2-bc}\right)\geq0$$ or $$\sum_{cyc}\frac{a^3(a^2+b^2+c^2-3bc)}{bc(a^2+b^2+c^2-bc)}\geq0$$ or $$\sum_{cyc}\frac{a^4(a^2+b^2+c^2-3bc)}{a^2+b^2+c^2-bc}\geq0.$$ But $$\left(\frac{a^4}{a^2+b^2+c^2-bc},\frac{b^4}{a^2+b^2+c^2-ac},\frac{c^4}{a^2+b^2+c^2-ab}\right)$$ and $$(a^2+b^2+c^2-3bc,a^2+b^2+c^2-3ac,a^2+b^2+c^2-3ab)$$ are the same ordered.

Thus, by Chebyshov we obtain $$\sum_{cyc}\frac{a^4(a^2+b^2+c^2-3bc)}{a^2+b^2+c^2-bc}\geq\frac{1}{3}\sum_{cyc}\frac{a^4}{a^2+b^2+c^2-bc}\sum_{cyc}(a^2+b^2+c^2-3bc)=$$ $$=\sum_{cyc}\frac{a^4}{a^2+b^2+c^2-bc}\sum_{cyc}(a^2-ab)\geq0.$$ Done!