# Prove: $\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\ge\sqrt{3}\sum_{cyc}{\sqrt[4]{\frac{5ab}{c}+4a}}$

Prove that the following inequality :$$\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\ge\sqrt{3}\left(\sqrt[4]{\frac{5ab}{c}+4a}+\sqrt[4]{\frac{5bc}{a}+4b}+\sqrt[4]{\frac{5ca}{b}+4c}\right)$$ holds for all positive real numbers such that: $$abc=1$$.

I saw problem on: AoPS. I tried to continue Arqady comment:

By Holder $$\sqrt3\sum_{cyc}\sqrt[4]{\frac{5ab}{c}+4a}=\sqrt3\sum_{cyc}\sqrt[4]{ab(5ab+4ac)}\leq$$ $$\leq\sqrt3\sqrt[4]{\left(\sum\limits_{cyc}\sqrt[4]{ab}\right)^3\sum_{cyc}\sqrt[4]{ab}(5ab+4ac)}\leq(\sqrt a+\sqrt b+\sqrt c)^2.$$ We have: $$\sqrt[4]{ab}+\sqrt[4]{bc}+\sqrt[4]{ca}\le\sqrt{a}+\sqrt{b}+\sqrt{c}$$

The rest is prove the following inequality: $$9\sum_{cyc}{\sqrt[4]{ab}(5ab+4ac)}\leq(\sqrt{a}+\sqrt{b}+\sqrt{c})^5$$ However, it is not true by acountable example. I hope we can find a good solution.

Thank you!

• But my way gives a right inequality, which theoretically we can prove without computer :) Oct 17, 2021 at 10:56
• I am sorry, but What do you mean "my way"? Oct 17, 2021 at 12:13
• In your post you showed my solution. Oct 17, 2021 at 19:18
• Are you Arqady on AoPS? Oct 18, 2021 at 1:40
• Yes, of course. Oct 18, 2021 at 3:54

We need to prove that $$(a+b+c)^2\geq\sum_{cyc}\sqrt[4]{9a^4b^2(5b^2+4c^2)}.$$ Now, by Holder and Rearrangement we obtain: $$\sum_{cyc}\sqrt[4]{9a^4b^2(5b^2+4c^2)}=\sum_{cyc}\sqrt[4]{9a^2b^2\cdot a(5b^2a+4c^2a)}\leq$$ $$\leq\sqrt[4]{9(ab+ac+bc)^2(a+b+c)\sum_{cyc}(5b^2a+4c^2a)}\leq$$ $$\leq\sqrt[4]{9(ab+ac+bc)^2(a+b+c)\left(4\sum_{cyc}(a^2b+a^2c)+\frac{4(a+b+c)^3}{27}-abc\right)}.$$ Now, let $$a+b+c=3u$$, $$ab+ac+bc=3v^2$$ and $$abc=w^3$$.
Thus, $$u^2\geq v^2$$ and we need to prove that: $$(3u)^7\geq9(3v^2)^2(4(9uv^2-3w^3)+4u^3-w^3)$$ or $$27u^7-4u^3v^4-36uv^6+13v^4w^3\geq0$$ and since by Schur $$w^3\geq4uv^2-3u^3,$$ it's enough to prove that: $$27u^7-4u^3v^4-36uv^6+13v^4(4uv^2-3u^3)\geq0$$ or $$u(u^2-v^2)(27u^4+27u^2v^2-16v^4)\geq0,$$ which is obvious.