Prove: $\sqrt{\frac{a}{4a^2+bc}}+\sqrt{\frac{b}{4b^2+ac}}+\sqrt{\frac{c}{4c^2+ba}}\ge \frac{\sqrt{2}}{\sqrt{a+b+c}}.$

Question

Problem: Let $a,b,c\ge0: ab+bc+ca>0$. Prove that: $$\sqrt{\frac{a}{4a^2+bc}}+\sqrt{\frac{b}{4b^2+ac}}+\sqrt{\frac{c}{4c^2+ba}}\ge \frac{\sqrt{2}}{\sqrt{a+b+c}}.$$

Here is my attempt:

My idea is using Holder inequality.WLOG, assume that $a+b+c=2$ and we prove $$\sqrt{\frac{a}{4a^2+bc}}+\sqrt{\frac{b}{4b^2+ac}}+\sqrt{\frac{c}{4c^2+ba}}\ge 1.$$

By Holder: $$\left(\sum\limits_{cyc}\dfrac{\sqrt{a}}{\sqrt{4a^2+bc}}\right)^2\left[\sum\limits_{cyc}a^2(b+c)^3(4a^2+bc)\right] \ge \left[\sum\limits_{cyc}a(b+c)\right]^3=8(ab+bc+ca)^3$$ It's enough to prove that: $$8(ab+bc+ca)^3 \ge a^2(b+c)^3(4a^2+bc)+b^2(a+c)^3(4b^2+ac)+c^2(b+a)^3(4c^2+ba).$$ But it is not true! For example, $b=c=0.3; a=1.4$ leads to wrong inequality. I hope we can find better idea. Thank you.

Reminds me of a question I asked years ago. Not exactly the same, but related: https://mathoverflow.net/questions/295114/how-to-prove-that-a-b-4-sqrt1-a2-b2-leq-4-sqrta2-b2 — Max Muller, Jun 13 '23 at 13:20

Michael Rozenberg · Answer 1 · 2023-06-13T14:31:44.740

1

For $abc=0$ your inequality is obviously true.

Let $abc\neq0.$

Thus, we need to prove that: $$\sum_{cyc}\sqrt{\frac{\frac{1}{a}}{\frac{4}{a^2}+\frac{1}{bc}}}\geq\sqrt{\frac{2}{\sum\limits_{cyc}\frac{1}{a}}}$$ or $$\sum_{cyc}\frac{1}{\sqrt{a^2+4bc}}\geq\sqrt{\frac{2}{ab+ac+bc}}.$$ Now, by Holder $$\sum_{cyc}\frac{1}{\sqrt{a^2+4bc}}=\sqrt{\frac{\left(\sum\limits_{cyc}\frac{1}{\sqrt{a^2+4bc}}\right)^2\sum\limits_{cyc}(a^2+4bc)(b+c)^3(2a+b+c)^3}{\sum\limits_{cyc}(a^2+4bc)(b+c)^3(2a+b+c)^3}}\geq$$ $$\geq\sqrt{\frac{\left(\sum\limits_{cyc}(b+c)(2a+b+c)\right)^3}{\sum\limits_{cyc}(a^2+4bc)(b+c)^3(2a+b+c)^3}}\geq\sqrt{\frac{2}{ab+ac+bc}},$$ where the last inequality it's $$4(ab+ac+bc)\left(\sum_{cyc}(a^2+3ab)\right)^3\geq\sum\limits_{cyc}(a^2+4bc)(b+c)^3(2a+b+c)^3$$ or $$\sum_{sym}(11a^6b^2+46a^5b^3+38a^4b^4+14a^6bc+210a^5b^2c+562a^4b^3c+495a^4b^2c^2+800a^3b^3c^2)\geq0$$ and we are done.

edited Jun 13 '23 at 14:31

answered Jun 13 '23 at 14:26

Michael Rozenberg

194,933

It is powerful, Michael Rozenberg. Thank you. In post #12: https://artofproblemsolving.com/community/c6t243f6h2781273_very_nice_abbcca1 I am inspired of yours. Do you think it is true ? – TATA box Jun 13 '23 at 14:52
@Trần Nk Trang Thank you! The step with $(2a+b+c)^3$ it's a hardest step. I'll try to prove your linked inequality. – Michael Rozenberg Jun 13 '23 at 15:27
The following is also true $$\sqrt{\frac{a}{3a^2+2bc}}+\sqrt{\frac{b}{3b^2+2ac}}+\sqrt{\frac{c}{3c^2+2ba}}\ge \frac{2\sqrt{6}}{3\sqrt{a+b+c}}.$$ – TATA box Jun 14 '23 at 04:15
@Trần Nk Trang Yes, we can prove it by the same way. – Michael Rozenberg Jun 14 '23 at 07:14
in linked problem, does uvw work ? – TATA box Jun 14 '23 at 07:32
@Trần Nk Trang Yes, works. But I am not sure that $b=a$ and $c=\frac{1-a^2}{2a}$ leads to a right inequality. – Michael Rozenberg Jun 14 '23 at 07:48
Let us continue this discussion in chat. – TATA box Jun 14 '23 at 07:55
Interestingly, $$\frac{a}{4a^2+bc}+\frac{b}{4b^2+ca}+\frac{c}{4c^2+ab}\ge\frac{1}{a+b+c}$$ also holds. – TATA box Jun 14 '23 at 09:47
Yes, but it's very easy. – Michael Rozenberg Jun 14 '23 at 10:01
I agree with you. The inequality is obviously true after full expanding. Does Holder work ? By the way, I think the following is also true: $$\sqrt[3]{\frac{a}{8a^2+bc}}+\sqrt[3]{\frac{b}{8b^2+ac}}+\sqrt[3]{\frac{c}{8c^2+ba}}\ge \sqrt[3]{\frac{2}{a+b+c}}.$$ – TATA box Jun 14 '23 at 10:26
Can down voter explain me, why did you do it? – Michael Rozenberg Jul 29 '23 at 04:37

Prove: $\sqrt{\frac{a}{4a^2+bc}}+\sqrt{\frac{b}{4b^2+ac}}+\sqrt{\frac{c}{4c^2+ba}}\ge \frac{\sqrt{2}}{\sqrt{a+b+c}}.$

1 Answers1

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