# Prove inequality $2\le\frac{a+b}{c+1} +\frac{b+c}{a+1} +\frac{c+a}{b+1} \le3$

Assume that $a,b,c$ are real numbers from the interval $(\frac{1}{2},1)$. What is the proof that

$$2\le\frac{a+b}{c+1} +\frac{b+c}{a+1} +\frac{c+a}{b+1} \le3$$ holds?

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What do you know? What did you try? How do you know the result holds? – Did Aug 27 '12 at 22:49

For the first inequality

Since the sum is symmetric, without loss of generality we can assume $a \leq b \leq c$.

Then

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

Then, by Chebyshev's sum inequality we have

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

Thus

$$\left( \frac{a+b}{c+1} +\frac{b+c}{a+1} +\frac{c+a}{b+1} \right) \geq \frac{6(a+b+c)}{a+b+c+3}$$

It is easy to see that $$\frac{6(a+b+c)}{a+b+c+3} \geq 2 \Leftrightarrow a+b+c \geq \frac{3}{2}$$

Second Inequality This is equivalent to

$$\sum (a+b)(a+1)(b+1) \leq 3 (a+1)(b+1)(c+1)$$

Again, without loss of generality we can assume that $a \leq b \leq c$.

Then

$$(a+b)(a+1)(b+1) =(a+b)(ab+a+b+1)=a^2b+ab^2+a^2+b^2+2ab+a+b$$

Thus, the inequality becomes

$$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b+2a^2+2b^2+2c^2+2ab+2ac+2bc+2a+2b+2c \leq$$ $$\leq 3abc+3ab+3ac+3bc+3a+3b+3c+3$$

Or

$$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b+2a^2+2b^2+2c^2 \leq$$ $$\leq 3abc+ab+ac+bc+a+b+c+3 (*)$$

We start by observing that $(a+1)(1-b)(1-c) \geq 0$ thus

$$abc+a+bc +1 \geq ab+ac+b+c$$ Similarly $$abc+b+ac +1 \geq bc+ab+a+b$$ $$abc+c+ab +1 \geq ac+bc+a+b$$

Adding them we get

$$ab+ac+bc+a+b+c \leq 3abc+3$$

Thus, to prove $(*)$ it is enough to show that

$$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b+2a^2+2b^2+2c^2 \leq$$ $$\leq 2(ab+ac+bc+a+b+c)$$

which follows immediately from $a\leq 1 \Rightarrow a^2b \leq ab$ and similar ones.

*P.S. * This Solution is terrible, there should be a much simpler one. The inequality $(a+1)(1-b)(1-c) \geq 0$ I used suggests that the Schur Inequality should be the Key, but I couldn't find it :)

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For the left inequality: Since the sum of any two of $a,b,c$ is greater than 1, one has that $a+1$, $b+1$, $c + 1 < a + b + c$. So you have $$\frac{a+b}{c+1} +\frac{b+c}{a+1} +\frac{c+a}{b+1} > \frac{a+b}{a + b + c} +\frac{b+c}{a + b + c} +\frac{c+a}{a + b + c}$$ $$= \frac{2a + 2b + 2c}{a + b + c}$$ $$= 2$$

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