# How prove this $\frac{a-b}{a+2b+c}+\frac{b-c}{b+2c+d}+\frac{c-d}{c+2d+e}+\frac{d-e}{d+2e+a}+\frac{e-a}{e+2a+b}\ge 0$

Let $a,b,c,d,e$ are postive real numbers,show that $$\dfrac{a-b}{a+2b+c}+\dfrac{b-c}{b+2c+d}+\dfrac{c-d}{c+2d+e}+\dfrac{d-e}{d+2e+a}+\dfrac{e-a}{e+2a+b}\ge 0$$

My try: since $$\Longleftrightarrow\sum_{sym}\left(\dfrac{a-b}{a+2b+c}+\dfrac{1}{2}\right)\ge\dfrac{5}{2}$$ $$\Longleftrightarrow \sum_{sym}\left(\dfrac{3a+c}{a+2b+c}\right)\ge 5$$ use Cauchy-Schwarz inequality,we have $$\sum_{sym}\dfrac{3a+c}{a+2b+c}\sum_{sym}((3a+c)(a+2b+c))\ge\left(\sum_{sym}(3a+c)\right)^2$$

$$\Longleftrightarrow 16(\sum_{sym}a)^2\ge5\sum_{sym}(3a+c)(a+2b+c)$$

becasue this is not hold $$\Longleftrightarrow 16(\sum_{sym}a)^2\ge5\sum_{sym}(3a+c)(a+2b+c)$$for $a,b,c,d,e>0$ and this method is from this simaler inequality : How prove this inequality $\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+e}+\frac{d-e}{e+a}+\frac{e-a}{a+b}\ge 0$ then I can't,Thank you

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Aren't you working with cyclic sums instead of symmetric sums? – Ragnar Dec 30 '13 at 13:49

You need to try a different fraction to add, so that you get better terms on applying Cauchy-Schwarz. For e.g. here we have the equivalent inequality:

$$\sum_{cyc} \left(\frac{a-b}{a+2b+c}+\frac15\right)\ge 1$$

$$\iff \sum_{cyc} \frac{6a-3b+c}{a+2b+c} \ge 5$$

By Cauchy-Schwarz, we have: $$\sum_{cyc} \frac{6a-3b+c}{a+2b+c} \ge \frac{\left(\sum_{cyc} (6a-3b+c)\right)^2}{\sum_{cyc}\left((a+2b+c)(6a-3b+c)\right)} = \frac{16(a+b+c+d+e)^2}{\sum_{cyc}(a^2+8ab+7ac)}$$

So it is sufficient if we can show that: $$16(a+b+c+d+e)^2 \ge 5 \sum_{cyc}(a^2+8ab+7ac)$$

But we can express the above $LHS - RHS$ as: $$4\sum_{cyc}(a-b)^2 + \frac32 \sum_{cyc}(a-c)^2 \ge 0$$

Added based on a comment in linked post - this application of Cauchy Schwarz requires the numerator of the fraction $6a-3b+c$ to be non-negative, and hence covers only the cases with a condition like $a,b,c,d,e \in [\frac1k,k]$ where $k^2=\frac73$.

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I'll solve the three variable case (the $a,b,c,d,e$ case is similar).

Assume without loss of generality $a \geq b \geq c$. Multiplying by the denominators we obtain

$$a^3+a^2 c+a b^2-6 a b c+b^3+b c^2+c^3 \geq 0$$

Now use the rearrangment inequality twice. We can split the inequality in two parts: $$a^2c+b^2a+c^2b \geq 3abc$$ and $$a^3+b^3+c^3 \geq 3abc$$ For the first one, use the rearrangment inequality to the sequences $ab, ac, bc$ and $a,b,c$ (which are listed in increasing order). As for the second one, you can use the extension of the rearrangement inequality (also explained in the link) with three equal sequences $a,b,c$. Sum up the two inequalities and you are done.

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This should work, but is a quick and dirty solution. My intuition tells me so, because it is "analogous", but this would need to generalize the coefficient after expanding, and that would go a little nasty... – chubakueno Dec 30 '13 at 15:22
Totally agree. But since nobody else posted anything... I can change the analogous for a "similar". – gerd Dec 30 '13 at 15:26
NO,Three variable is very easy,becasue can use Cauchy-Schwarz inequality to kill it,But for five variable inequality can't it,and you methods will very ugly! – math110 Dec 30 '13 at 15:35
Try induction using the rearrangement inequality. Obviously you wouldn't expand it in the 5,6,7... case, but you would prove it in the $n$th one. – gerd Dec 30 '13 at 15:38
@math110 Please be polite. Also, having a better grammar wouldn't do any harm :) – chubakueno Dec 30 '13 at 16:16