# Proving : $\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}\ge1$

Let $a,b,c > 0$ be a real numbers ,such that : $abc=1$,how to prove that:

$$\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}\ge1$$

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should we create a symmetric-inequalities tag? – leonbloy Jul 4 '12 at 17:03
@leonbloy I am no in favor of the idea – Belgi Jul 4 '12 at 18:02

Since the inequality is symmetric in $a,b,c$ without loss of generality we can assume that $a \geq b \geq c$. Then

$$b+c+1 \leq a+c+1 \leq a+b+1$$ and $$\frac{a}{b+c+1} \geq \frac{b}{c+a+1} \geq \frac{c}{a+b+1}$$

Now, by Chebyshev's sum inequality you have

$$\frac{1}{3} \left[ \frac{a}{b+c+1} \cdot (b+c+1)+\frac{b}{c+a+1}\cdot (a+c+1)+\frac{c}{a+b+1}\cdot (a+b+1) \right]$$ $$\leq \frac{1}{9}\left[ \frac{a}{b+c+1} +\frac{b}{c+a+1}+\frac{c}{a+b+1}\right]\left[ (b+c+1)+ (a+c+1)+ (a+b+1) \right]$$

or equivalently

$$3(a+b+c) \leq \left[ \frac{a}{b+c+1} +\frac{b}{c+a+1}+\frac{c}{a+b+1}\right]\left[ 2a+2b+2c+3 \right] \,. \tag{*}$$

Now, by AM-GM you have

$$1 \leq \sqrt[3]{abc} \leq \frac{a+b+c}{3}$$

hence

$$2a+2b+2c+3 \leq 3(a+b+c) \tag{*\!*}$$

Combining $(*)$ with $(**)$ you get your desired inequality.

P.S. I don't know why, probably experience with these, but when I saw it the inequality screamed Chebyshev to me...

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Just a note: you can use \tag{something} to name your equations. It places them at the far alligned right, so it doesn't get mixed up as part of the actual equation. Other then that, nice post! – Joe Jul 4 '12 at 16:08
@Frank: A bit late to comment, but the solution isn't quite right. If you have $a \ge k$, it doesn't imply $\dfrac{1}{a} \ge \dfrac{1}{k}$. – Inceptio Jun 20 '13 at 13:48

As A.M.≥G.M. for positive real numbers, $a+b+c ≥ 3(abc)^{1/3}$ = 3

L.H.S. = $\sum\frac{a}{b+c+1}$ = -3+$\sum(\frac{a}{b+c+1}$+1) = -3 + (a+b+c+1)$\sum\frac{1}{b+c+1}$

As A.M.≥H.M. for positive real numbers, clearly $\frac{1}{b+c+1}$>0.

So, (1/3)$\sum\frac{1}{b+c+1} ≥ 3/\sum(b+c+1)$ taking A.M. and H.M. of $\frac{1}{b+c+1}$ etc.,

Or, $\sum(b+c+1) \sum\frac{1}{b+c+1}$ ≥ 9,

Or, (2(a+b+c)+3) $\sum\frac{1}{b+c+1}$ ≥ 9,

Let a+b+c=3d where d≥1 => 2(a+b+c)+3=6d+3 => $\sum\frac{1}{b+c+1}≥\frac{9}{6d+3}=\frac{3}{2d+1}$

L.H.S. = -3 + (3d+1) $\frac{3}{2d+1}=\frac{9d+3}{2d+1} - 3 =\frac{3d}{2d+1}$

Now, $\frac{3d}{2d+1}$ will be ≥1 if 3d≥2d+1 or if d≥1 which is true.

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You have some severe formatting problems, e.g. non-TeX math, unmatched parentheses and undefined symbol Σ (is this a sum or a sigma? this is why I didn't edit the post myself). For math please use $\TeX$, for example $\frac{a}{b+c+1}$ can be typeset as $\frac{a}{b+c+1}$, sums $\sum_{a}^{b}c$ $\sum_{a}^{b}c$, sigma-s $\Sigma$, $\Sigma$, implications $\Rightarrow$ $\Rightarrow$, inequalities $\leq \geq$ $\leq \geq$. – dtldarek Jul 4 '12 at 18:17
@lab bhattacharjee: How does $(2(a+b+c)+3)\sum\frac{1}{b+c+1}\geq 9$ and $2(a+b+c)+3\geq 9$ imply $\sum\frac{1}{b+c+1}\geq 1$ ? You are saying that $xy\geq 9$ and $x\geq 9$ imply $y\geq 1$ which is not true, take $x=18,y=1/2$ – pritam Jul 5 '12 at 14:56
Thanks Pritam for your observation, would you please verify once more? – lab bhattacharjee Jul 5 '12 at 17:47
@labbhattacharjee: Yeah now it looks correct. – pritam Jul 5 '12 at 18:05

As A.M.≥H.M. for positive real numbers, clearly 1b+c+1>0.

So, (1/3)∑1b+c+1≥3/∑(b+c+1) taking A.M. and H.M. of 1b+c+1 etc.,

Or, ∑(b+c+1)∑1b+c+1 ≥ 9,

Or, (2(a+b+c)+3) ∑1b+c+1 ≥ 9,

Let a+b+c=3d where d≥1 => 2(a+b+c)+3=6d+3 => ∑1b+c+1≥96d+3=32d+1 L.H.S. = -3 + (3d+1) 32d+1=9d+32d+1−3=3d2d+1

Now, 3d2d+1 will be ≥1 if 3d≥2d+1 or if d≥1 which is true.