Take the 2-minute tour ×
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.

Let $a,b,c$ be positive numbers. Prove that $$\dfrac{(b+c-a)^2}{(b+c)^2+a^2}+ \frac{(c+a-b)^2}{(c+a)^2+b^2}+ \frac{(a+b-c)^2}{(a+b)^2+c^2} \ge \frac{3}{5}$$

share|improve this question
    
Are you the same person? If so, I can flag the moderator to merge your accounts. Kindly use a single account. –  user17762 Jan 20 '13 at 3:46
add comment

1 Answer

up vote 7 down vote accepted

You labeled this as homework, but I am not sure what methods you have learnt. (Are you in HSGS of Vietnam for example?) I would assume that you know the tangent line method, and will add more explanation if you need it.

Since LHS is homogeneous, we may assume that $a+b+c = 3$. So we need to prove that $$\frac{(3-2a)^2}{(3-a)^2+a^2} + \frac{(3-2b)^2}{(3-b)^2+b^2} + \frac{(3-2c)^2}{(3-c)^2+c^2} \ge \frac{3}{5}$$ Note that $$\frac{(3-2a)^2}{(3-a)^2+a^2} \ge \frac{1}{5} - \frac{18}{25}(a-1)$$ (This is equivalent to $(2a+1)(a-1)^2 \ge 0$) So $$\frac{(3-2a)^2}{(3-a)^2+a^2} + \frac{(3-2b)^2}{(3-b)^2+b^2} + \frac{(3-2c)^2}{(3-c)^2+c^2} \ge \frac{3}{5} - \frac{18}{25}(a-1+b-1+c-1) = \frac{3}{5}$$

share|improve this answer
    
Neat. +1. Could you add more details for the step? $$\frac{(3-2a)^2}{(3-a)^2+a^2} \ge \frac{1}{5} - \frac{18}{25}(a-1)$$ –  user17762 Jan 20 '13 at 4:02
    
@Marvis, You can check by clearing the denominator that it is equivalent to $(2a+1)(a-1)^2 \ge 0$. As for the motivation, let $f(a) = \frac{(3-2a)^2}{(3-a)^2+a^2}$. Then $f(1) = \frac{1}{5}$, and $f'(1) = -\frac{18}{25}$. –  user27126 Jan 20 '13 at 4:04
    
Why you can assume $a+b+c=3$ ? –  harrypham Jan 20 '13 at 4:18
    
@harrypham, if not, replace $a$ by $\frac{3a}{a+b+c}$ and so forth. Left hand side doesn't change this way. –  user27126 Jan 20 '13 at 4:19
add comment

Your Answer

 
discard

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.