Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

(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 :)

share|cite|improve this question
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
up vote 1 down vote accepted

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:))

share|cite|improve this answer
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 zh Sep 13 '12 at 1:01

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

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


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

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.