Prove that $\frac{a}{(1+b)(1+c)}+\frac{b}{(1+a)(1+c)}+\frac{c}{(1+a)(1+b)} \geq \frac{3}{4}.$

Question

Let $a,b,c$ be positive numbers that satisfy $abc = 1$, prove that $$\dfrac{a}{(1+b)(1+c)}+\dfrac{b}{(1+a)(1+c)}+\dfrac{c}{(1+a)(1+b)} \geq \dfrac{3}{4}.$$

Attempt

I tried doing $$\dfrac{a}{(1+b)(1+c)}+\dfrac{b}{(1+a)(1+c)}+\dfrac{c}{(1+a)(1+b)} = \dfrac{1}{bc(1+b)(1+c)}+\dfrac{1}{ac(1+a)(1+c)}+\dfrac{c}{bc(1+a)(1+b)}$$ then using the fact that $a^2+b^2+c^2 \geq ab+bc+ca$ but that only seemed to get me $$\dfrac{a}{(1+b)(1+c)}+\dfrac{b}{(1+a)(1+c)}+\dfrac{c}{(1+a)(1+b)} \geq \dfrac{1}{bc+bc^2+c+1}+\dfrac{1}{ac+a^2c+c+1}+\dfrac{1}{ab+ab^2+b+1}.$$

Use Cauchy inequality. Just sum it and use Cauchy. – openspace Dec 31 '15 at 03:00 — openspace, Dec 31 '15 at 03:00

score 3 · Answer 1 · answered Dec 31 '15 at 03:09

3

The inequality is equivalent to $$4(a(1+a)+b(1+b)+c(1+c))\ge 3(1+a)(1+b)(1+c).\tag{1}$$ Since $$(1+a)(1+b)(1+c) = 1+ (a+b+c)+(ab+bc+ca)+abc$$ and $abc=1$, Inequality (1) is equivalent to $$4(a^2+b^2+c^2)+(a+b+c) \ge 6+3(ab+bc+ca).$$ The last inequality follows from $a^2+b^2+c^2\ge ab+bc+ca$ and AM-GM.

answered Dec 31 '15 at 03:09

Quang Hoang

15,854

Can you elaborate on the last part? – Jacob Willis Dec 31 '15 at 03:13
You beat me to it I suppose... – Panglossian Oporopolist Dec 31 '15 at 03:13
Why is $4(a^2+b^2+c^2)+(a+b+c) \ge 6+3(ab+bc+ca)$? – Jacob Willis Dec 31 '15 at 03:15
$4(a^2+b^2+c^2)+(a+b+c) \ge 3(ab+bc+ca)+(a^2+b^2+c^2)+(a+b+c)\ge 3(abc)^\frac13+3(abc)^\frac23=6+3(ab+bc+ca)$ by AM–GM – GNUSupporter 8964民主女神地下教會 Dec 31 '15 at 03:24
I'm sorry I still don't follow. – Jacob Willis Dec 31 '15 at 03:26
@JacobWillis $$\frac{a^2+b^2+c^2}{3}\ge\sqrt[3]{a^2b^2c^2}=1$$ $$\frac{a+b+c}{3}\ge\sqrt[3]{abc}=1$$ – GNUSupporter 8964民主女神地下教會 Dec 31 '15 at 03:33

openspace · Answer 2 · 2015-12-31T06:34:48.087

1

Ok, let's sum it: $$\frac{(a+1)a+b(b+1)+c(c+1)}{(a+1)(b+1)(c+1)}$$, called it $(1)$

Now use Cauchy inequality: $$(1) >= \frac{3((a+1)ab(b+1)c(c+1))^{\frac{1}{3}}}{(a+1)(b+1)(c+1)}$$, now

Derived all, what we can derive :)

And use Cauchy for $a+1,b+1,c+1$.

So continue it!

Hint: show that: $$((a+1)(b+1)(c+1))^{\frac{2}{3}} >= 4$$

edited Dec 31 '15 at 06:34

answered Dec 31 '15 at 03:11

openspace

6,470

I posted it from my phone, so it's hard to write full solution. – openspace Dec 31 '15 at 03:14
Your inequality got flipped at some point. – Quang Hoang Dec 31 '15 at 05:55
@QuangHoang are you talking about $>=$ in forth line ? – openspace Dec 31 '15 at 05:57
Your 2nd inequality is $(1)>3 \left( (a+1)(b+1)(c+1)\right)^{\color{red}-2/3}$. The RHS of which is less than $1/4$. – Quang Hoang Dec 31 '15 at 06:14
@QuangHoang yep – openspace Dec 31 '15 at 06:34

score 0 · Answer 3 · answered Dec 31 '15 at 03:13

Hint: Multiply both sides by $(1+a)(1+b)(1+c)$, and expand the aforementioned expression as follows: $1+abc+a+b+c+ab+ac+bc$.

You'll notice that you can substitute the value $1$ for $abc$; I think you should be able to tackle it from here by application of RM-AM-GM-HM inequality (Only AM-GM necessary though).

Prove that $\frac{a}{(1+b)(1+c)}+\frac{b}{(1+a)(1+c)}+\frac{c}{(1+a)(1+b)} \geq \frac{3}{4}.$

3 Answers3