Prove $\sqrt{2a+b^3}+\sqrt{2b+c^3}+\sqrt{2c+a^3}\le 3\sqrt{3}$ for $a^2+b^2+c^2=3.$

Question

Problem. Given non-negative real numbers $a,b,c$ satisfying $a^2+b^2+c^2=3.$ Prove that $$\sqrt{2a+b^3}+\sqrt{2b+c^3}+\sqrt{2c+a^3}\le 3\sqrt{3}.$$ Naturally by using Cauchy-Schwarz, we need to show that$$2(a+b+c)+a^3+b^3+c^3\le 9,$$ which is not true when $ a=b\rightarrow 1^{-}.$

I also tried$$\sum_{cyc}\sqrt{2a+b^3}=\sum_{cyc}\sqrt{\frac{2a+b^3}{2a^2+b^2}\cdot (2a^2+b^2)}\le \sqrt{\sum_{cyc}\frac{2a+b^3}{2a^2+b^2}\cdot 3(a^2+b^2+c^2),}$$which saves occuring equality but $$\frac{2a+b^3}{2a^2+b^2}+\frac{2b+c^3}{2b^2+c^2}+\frac{2c+a^3}{2c^2+a^2}\le 3$$ is already wrong when $a=b=0.99.$

I hope you give some better approach using Holder, AM-GM, etc. Maybe the BW helps for the rest.

Thank you for paying attention.

Answer 1

Some thoughts.

Fact 1. Let $x, y, z \ge 0$ with $x^2 + y^2 + z^2 + \frac56x^2y^2z^2 \le \frac{23}{6}$. Then $x + y + z \le 3$. (It is verified by Mathematica. Similar to: here.)

Now, let $$x := \sqrt{\frac{2a + b^3}{3}}, \quad y := \sqrt{\frac{2b + c^3}{3}}, \quad z := \sqrt{\frac{2c + a^3}{3}}.$$ We have $$x^2 + y^2 + z^2 + \frac56x^2y^2z^2 \le \frac{23}{6}$$ which is equivalent to $$621 - 54(a^3 + b^3 + c^3 + 2a + 2b + 2c) - 5(b^3 + 2a)(c^3 + 2b)(a^3 + 2c)\ge 0. \tag{1}$$ (1) is verified by Mathematica. Is there a nice proof?

By Fact 1, we have $x + y + z \le 3$.

Answer 2

By C-S $$\sum_{cyc}\sqrt{2a+b^3}\leq\sqrt{\sum_{cyc}\frac{2a+b^3}{3a+5b+c}\sum_{cyc}(3a+5b+c)}=3\sqrt{(a+b+c)\sum_{cyc}\frac{2a+b^3}{3a+5b+c}}$$ and it's enough to prove that: $$(a+b+c)\sum_{cyc}\frac{2a+b^3}{3a+5b+c}\leq3,$$ which is true, but my solution is not so good: homogenization and B-W.

About BW see here: https://artofproblemsolving.com/community/c6h522084p2942645