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.

If we assume $S = \sqrt{2a+b^3} + \sqrt{2b+c^3} + \sqrt{2c+a^3}$ then using Cauchy-Schwarz ineqaulity we get $S^2 \leq 3(a^3+b^3+c^3 + 2(a+b+c))$ then using $AM-GM$ we get $S^2 \leq 3(2(a+b+c)+3abc)$ Now I am also stuck :( — Lucky Chouhan, Nov 23 '23 at 15:21

Michael Rozenberg · Answer 1 · 2023-11-24T03:57:34.423

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

edited Nov 24 '23 at 03:57

answered Nov 23 '23 at 18:09

Michael Rozenberg

194,933

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You used Bacteria skill. Thank you. – TATA box Nov 23 '23 at 23:36
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What is B-W and how can we complete the proof? – Sickness Nov 24 '23 at 01:35

score 1 · Accepted Answer · answered Nov 25 '23 at 00:33

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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 $x + y + z \le 3$.

answered Nov 25 '23 at 00:33

River Li

37,323

1

Nice idea. I also to try: Let $x, y, z \ge 0$ with $x^2 + y^2 + z^2 + \frac34x^2y^2z^2 \le \frac{15}{4}$. Then $x + y + z \le 3$ but it failed. – TATA box Nov 25 '23 at 00:36
By the way, is Mathematica absolutely correct in inequalities? I think it might make a mistake. – youthdoo Nov 25 '23 at 04:06
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@youthdoo I use Mathematica to verify the inequality. This is not the end. At this stage, I will write "some thoughts". Then I will find a human verifiable proof. This time, I write "proof". – River Li Nov 25 '23 at 05:21
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@TATAbox You should not accept my answer since I don't have a nice proof of (1). I think Michael Rozenberg's better (already upvoted before I posted my answer). – River Li Nov 25 '23 at 10:10
I think his final step is hard. – TATA box Nov 25 '23 at 10:43
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@TATAbox Anyway, hope there is a nice proof of (1). – River Li Nov 25 '23 at 11:06
Did you use BW? – TATA box Nov 25 '23 at 11:35
@TATAbox I would edit it in some days. – River Li Nov 25 '23 at 11:54
I tried $(1)$. It is complicated. – TATA box Nov 28 '23 at 09:22
@TATAbox Yes, no real progress. – River Li Nov 28 '23 at 09:26

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

2 Answers2