Prove that $\frac{a}{b^2 + c}+\frac{b}{c^2 + a}+\frac{c}{a^2 + b} \ge \frac{3}{2}$ when $a+b+c=3$

Question

I've tried solving Pham Kim Hung's famous inequality problem which he posted on aops in April 2007 for 2 days now. This is the problem: Show that $$\frac{a}{b^2 + c}+\frac{b}{c^2 + a}+\frac{c}{a^2 + b} \ge \frac{3}{2} \tag{*}$$ when $a, b, c > 0$ and $a+b+c=3$.

I've tried with many theorems and tricks (like AM-GM, C-S, etc.) but haven't really figured out a way.

The closest I could get was by solving a similar problem like this.

So, if you know the proof of (*) please share here. I'd be much thankful to you.

if fact, there's generalisation to this,$ \frac {a}{\sqrt {kb + c^2}} + \frac {b}{\sqrt {kc + a^2}} + \frac {c}{\sqrt {ka + b^2}} \ge \frac {3}{\sqrt {k + 1}}$ (when a+b+c=3) — StackpackedKar, Nov 24 '21 at 04:06
you can view that here: https://artofproblemsolving.com/community/c6h300129 — StackpackedKar, Nov 24 '21 at 04:07
Equality is achieved when $a=b=c=1$, which should probably give a hint. — Prometheus, Nov 24 '21 at 04:10
You must add the condition $a,b,c$ positive because for example if $(x,y,z)=(1,-2,4)$ then $LHS\lt\dfrac32$ — Piquito, Nov 24 '21 at 05:26
This has been asked and answered before: https://math.stackexchange.com/q/703518/42969, https://math.stackexchange.com/q/4149973/42969 — Martin R, Nov 24 '21 at 05:28
@MartinR The first one is quite different. In the 1st one, the usual (obvious) Cauchy-Bunyakovsky-Schwarz $(\sum_{\mathrm{cyc}} \frac{a}{b^2 + c}) (\sum_{\mathrm{cyc}} a(b^2 + c)) \ge (a + b + c)^2$ works. However, it does not work here. — River Li, Nov 24 '21 at 05:51
@MartinR The 2nd one is the same as this topic. If this topic is not closed, I will give another proof different from Michael Rozenberg's. — River Li, Nov 24 '21 at 05:54

StackpackedKar · Accepted Answer · 2021-11-24T06:30:23.837

4

$\sum_{cyc}\frac{a}{b+c^2}=\sum_{cyc}\frac{a^4}{a^3b+a^3c^2}\geq\frac{(a^2+b^2+c^2)^2}{\frac{1}{3}(a^2+b^2+c^2)^2+\frac{1}{3}(a^2+b^2+c^2)^2}=\frac{3}{2}$
(For numerator we use Titu's Lemma and simplify denominator :) )

Thanks to arqady
For more information vist https://artofproblemsolving.com/community/c6h401290p2234611

edited Nov 24 '21 at 06:30

answered Nov 24 '21 at 06:03

StackpackedKar

303

But it is arqady's proof. (+1) Anyway. By the way, I use the Cauchy-Bunyakovsky-Schwarz in the same way but the remaining is different from arqady's. – River Li Nov 24 '21 at 06:09
Thanks :). I guess he is MSE too. – StackpackedKar Nov 24 '21 at 06:13
@RiverLi You seem to have a beautiful inequality in your profile ;) – StackpackedKar Nov 24 '21 at 06:15
1

In MSE, he is Michael Rozenberg. You may correct the spelling aquardy (arqady). – River Li Nov 24 '21 at 06:16
Thanks. It is No. 3 of Xuezhi Yang's 22 conjectures in inequalities (Mr. Yang is a Chinese scholar). – River Li Nov 24 '21 at 06:20
@RiverLi Out of curiosity, do you have a link to this list of 22 conjectures about inequalities? I am also amazed by the inequality in your profile description. – Prism Dec 23 '22 at 10:30
@Prism https://www.doc88.com/p-4405905284406.html (Chinese). OR https://www.docin.com/p-456874670.html (Chinese). – River Li Dec 23 '22 at 11:38
@Prism You are welcome. – River Li Dec 24 '22 at 13:34

Prove that $\frac{a}{b^2 + c}+\frac{b}{c^2 + a}+\frac{c}{a^2 + b} \ge \frac{3}{2}$ when $a+b+c=3$

1 Answers1