Let $a, b, c$ be positive real numbers. Prove the inequality.

Question

Let $a, b, c$ be positive real numbers. Prove that $$\dfrac{2a^2b^2c^2}{a^3b^3+b^3c^3+c^3a^3}+\dfrac{1}{3} \geq \dfrac{3abc}{a^3+b^3+c^3}.$$ My solutions is: $$\dfrac{3abc}{a^3+b^3+c^3}-\dfrac{1}{3}-\dfrac{2a^2b^2c^2}{a^3b^3+b^3c^3+c^3a^3} \leq 0.$$ As $$a^3+b^3+c^3 \geq 3abc,$$ then $$\dfrac{1}{a^3+b^3+c^3} \leq \dfrac{1}{3abc}.$$ So we have $$\dfrac{3abc}{a^3+b^3+c^3}-\dfrac{1}{3}-\dfrac{2a^2b^2c^2}{a^3b^3+b^3c^3+c^3a^3} \leq \dfrac{3abc}{3abc}-\dfrac{1}{3}-\dfrac{2a^2b^2c^2}{3a^2b^2c^2}=1-\dfrac{1}{3}-\dfrac{2}{3} \leq 0.$$ I still doubt the correctness of my solution, so any help would be greatly appreciated. Thanks a lot!

Answer 1

Let $x^3+y^3+z^3=3u$, $x^3y^3+x^3z^3+y^3z^3=3v^2$,$x^3y^3z^3=w^3$ and $u^2=tv^2$.

Thus, $t\geq1$ and we need to prove that $$\frac{2w^2}{v^2}+1\geq\frac{3w}{u}$$ or $$2uw^2-3v^2w+uv^2\geq0.$$ Now, if $$9v^4-8u^2v^2\leq0,$$ so the inequality is obviously true.

Let $9v^4-8u^2v^2\geq0$ or $t\leq\frac{9}{8}.$

Thus, it's enough to prove that: $$w\geq\frac{3v^2+\sqrt{9v^4-8u^2v^2}}{4u}$$ or $$w^3\geq\left(\frac{3v^2+\sqrt{9v^4-8u^2v^2}}{4u}\right)^3.$$ But by Schur $$\sum_{cyc}(a^9-a^6b^3-a^6c^3+a^3b^3c^3)\geq0$$ or $$27u^3-27uv^2+3w^3-9uv^2+3w^3+3w^3\geq0,$$ which gives $$w^3\geq4uv^2-3u^3$$ and it's enough to prove that $$64u^3(4uv^2-3u^3)\geq\left(3v^2+\sqrt{9v^4-8u^2v^2}\right)^3$$ or $$64t^2(4-3t)\geq(3+\sqrt{9-8t})^3,$$ where $1\leq t\leq\frac{9}{8}.$

Now, let $\sqrt{9-8t}=p$.

Thus, $0\leq p\leq 1$, $t=\frac{9-p^2}{8}$ and we need to prove that: $$(9-p^2)^2\left(4-\frac{3(9-p^2}{8}\right)\geq(3+p)^3$$ or $$(3-p)^2(5+3p^2)\geq8(3+p)$$ or $$(1-p)(21-17p+15p^2-3p^3)\geq0$$ and since $$21-17p+15p^2-3p^3\geq12-24p+12p^2=12(1-p)^2\geq0,$$ we are done.