# Inequality. $\frac{(b+c)^2}{a^2+bc}+\frac{(c+a)^2}{b^2+ca}+\frac{(a+b)^2}{c^2+ab} \geq 6$

(Crux Mathematicorum) If $a,b,c$ are the sidelengths of a triangle, then prove the inequality:

$$\frac{(b+c)^2}{a^2+bc}+\frac{(c+a)^2}{b^2+ca}+\frac{(a+b)^2}{c^2+ab} \geq 6 .$$

Thanks :)

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I suppose the last denominator is $c^2 + ab$. Am I correct? –  Andrea Orta Sep 12 '12 at 10:39
you have asked so many inequality questions. –  noname1014 Sep 12 '12 at 13:25

The result follows from the method of Example 3.2.3. and Example 3.2.4. in this link (Without loss of generality,if $a\geq b\geq c$,then $a^2+bc\geq b^2+ca \geq c^2+ab$ and $a+b\geq a+c \geq b+c$.One of these inequalities relies on the condition "a,b,c are the sidelengths of a triangle",try it yourself:))

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how do you find this pdf file?it is hard for me to search by google. –  noname1014 Sep 12 '12 at 17:08
@Tao Hong 洪涛:The keywords are "cyclic inequality" and "olympiad" –  y zhao Sep 13 '12 at 1:01

By homogeneity, we may assume $a+b+c=1$. By CS:

$$\displaystyle \sum_i \frac{x_i^2}{y_i} \geq \frac{\left(\sum_i x_i\right)^2}{\sum_i y_i}$$

Therefore:

$$\displaystyle LHS \geq \frac{\left(2(a+b+c)\right)^2}{(a+b+c)^2 - (ab+bc+ca)} = \frac{4}{1-t}$$

where $t:=ab+bc+ca$. Thus, we should show that $t \geq \tfrac13$ to complete the problem. Unfortunately, by CS, we have that:

$$\displaystyle 2t = 1 - (a^2+b^2+c^2) \leq 1 - \tfrac13(a+b+c)^3 = \tfrac23$$

and so this bound falls below.

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For $a = 1/2, b = c = 1/4$ we have $a + b + c = 1$ and $ab + bc + ca = 5/16 < 1/3$. –  AlbertH Sep 12 '12 at 16:15
Yes, this is consistent with my comment. The CS gives too low a bound. y zhao's comment above uses Chebychev $(a/x+b/y+c/z) \geq 3(a+b+c)/(x+y+z)$ which leads to the solution. –  cbyn Sep 12 '12 at 20:20