How to prove $\sum \sqrt{\frac{a^2+2bc}{b+c}} \geq \frac{3\sqrt{2}}{2}\sqrt{a+b+c}$ for all $a,b,c \ge 0$?

Question

Let $a,b,c$ be non-negative real numbers such that $ab+bc+ca>0$, prove that $$\sqrt{\frac{a^2+2bc}{b+c}}+\sqrt{\frac{b^2+2ca}{c+a}}+\sqrt{\frac{c^2+2ab}{a+b}} \ge \frac{3\sqrt{2}}{2}\sqrt{a+b+c}$$

I found it here and also here. But there are no simple nice solutions at all, and I don't know how they can find the Holder's yields. Here's my motivation I try to use Cauchy Schwarz $$\left(\sum_{cyc} \sqrt{\frac{a^2+2bc}{b+c}}\right)^2 \cdot \sum_{cyc} (b+c)(2a^2+bc)^2 \ge (a+b+c)^6$$ but the equality cases $a=b=c$ or $(0,t,t)$ aren't ensured. So far, I haven't had any more ideas. Can someone help me with this problem? And tell me why people can find the Holder's yields. Thanks a lot!

score 1 · Answer 1 · answered Jan 10 '24 at 14:59

1

Some thoughts.

It suffices to prove that $$\sum_{\mathrm{cyc}} \sqrt{\frac{a^2 + 2bc}{b + c}\cdot \frac{2}{a + b + c}} \ge 3.$$

Fact 1. Let $x, y, z \ge 0$ with $x^2y^2+y^2z^2+z^2x^2 + \frac{33}{16}x^2y^2z^2 \ge \frac{81}{16}$. Then $x + y + z \ge 3$.
(Note: It is verified by Mathematica. A human verifiable proof is needed.)

Now, let $$x := \sqrt{\frac{a^2 + 2bc}{b + c}\cdot \frac{2}{a + b + c}}, \quad \mathrm{etc.}$$

We have $$x^2y^2+y^2z^2+z^2x^2 + \frac{33}{16}x^2y^2z^2 \ge \frac{81}{16}.$$ (This can be proved by the Buffalo Way (BW).)

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

answered Jan 10 '24 at 14:59

River Li

37,323

Wow, that's an interesting and powerful lemma! The rest is not hard, and I could solve it by myself. Thank you, River Li – 30 Anh Ti 711 Jan 11 '24 at 12:34
@30AnhTi711 I want to see a nice proof, although I can not find a nice proof. – River Li Jan 11 '24 at 12:36
The inequality is strong, I think. Hope someone will have nice proof for it. I also find this https://artofproblemsolving.com/community/u1051050h3231885p29609849 – 30 Anh Ti 711 Jan 11 '24 at 13:24
For this lemma and in case the OP, we can choose the range $33/16 M\le 5/2$ specificlly. – Hello world Mar 05 '24 at 16:56
1

@Helloworld Really? I am not sure. – River Li Mar 06 '24 at 01:15
For $5/2$ is is less calculates – Hello world Mar 07 '24 at 04:39
@Helloworld Right. I see. – River Li Mar 07 '24 at 05:20
@RiverLi Could you please help me prove this problem? $$27(a^4+b^4+c^4+d^4)+148abcd\ge (a+b+c+d)^4, \forall a,b,c,d\ge 0.$$ – Hello world Mar 07 '24 at 09:22
@Helloworld BW works. Also, let $d=1$ and try pqr. – River Li Mar 07 '24 at 09:58
@RiverLi How can we let $d=1$ ? – Hello world Mar 07 '24 at 10:29
@Helloworld The inequality is homogeneous. Or, you divide both sides by $d^4$. – River Li Mar 07 '24 at 11:03
1

@RiverLi Wow, it is nice. Thank you – Hello world Mar 07 '24 at 11:05

score 1 · Answer 2 · edited Jan 16 '24 at 06:12

1

From River Li's idea and the help of Khuong Trang, I can solve my problem. Choose $a+b+c=2$. Now we just need to prove $$\sum_{cyc} \frac{(a^2+2bc)(b^2+2ca)}{(c+a)(c+b)} + \frac{5(a^2+2bc)(b^2+2ca)(c^2+2ab)}{2(a+b)(b+c)(c+a)} \ge \frac{11}{2}$$ or after full expanding and $pqr$, we need to prove $$f(r)=135r^2+r(171-162q)+10q(q-1)^2 \ge 0$$ we have $$f'(r)=270r+171-162q$$

If $4/3 >q>23/26$, using Schur $$f(r) \ge 270\cdot \frac{8(q-1)}{9} +171-162q=3(26q-23) \ge 0$$

If $q \le 23/26$, then $270r+171-162q >270r+171-162\cdot \dfrac{23}{26}>0$ then $f(r)$ increase. Using Schur $$f(r) \ge f\left(\frac{8(q-1)}{9}\right)=10(q-1)(q-17/5)(q-4/3)\ge 0$$ always true for $4/3>q>1$. The last case $q<1$ easy to check $$135r^2+r(171-162q)+10q(q-1)^2 \ge 0$$ Hence, the proof is done.

edited Jan 16 '24 at 06:12

D.W.

4,540

answered Jan 16 '24 at 06:11

30 Anh Ti 711

137

Can I ask you, how did you get rid of radicals and get the very first inequality? The rest looks clear. – Aig Feb 12 '24 at 18:04
@Aig, I get it from River Li. I have no idea how to find it. It seems magical to cancel the radical. – 30 Anh Ti 711 Feb 13 '24 at 02:58
@RiverLi Please help, where does the initial inequality in this answer come from? – Aig Feb 13 '24 at 04:14
@Aig I didn't see your comment. Just like pqr, we seek some relations between $x^2, y^2, z^2$ which implies the desired inequality. – River Li Mar 06 '24 at 01:17
@RiverLi Does isolated fudging help in this case? – Hello world Mar 07 '24 at 09:22
@Helloworld I didn't try. – River Li Mar 07 '24 at 09:51
@RiverLi Liu Quan Bao tried without success. – Hello world Mar 07 '24 at 10:29
@Helloworld Perhaps it is difficult to find the appropriate form. – River Li Mar 07 '24 at 11:06
@RiverLi Have you visit this problem: https://math.stackexchange.com/questions/4874800/prove-frac1-sqrtab2-frac1-sqrtbc2-frac1-sqrtca2-ge I think it is very nice and hard. – Hello world Mar 07 '24 at 16:18
@Helloworld I don't have a nice proof. – River Li Mar 08 '24 at 01:22

How to prove $\sum \sqrt{\frac{a^2+2bc}{b+c}} \geq \frac{3\sqrt{2}}{2}\sqrt{a+b+c}$ for all $a,b,c \ge 0$?

2 Answers2