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.

I recently met the inequality $\frac{a+b-2c}{b+c} + \frac{b+c-2a}{c+a} + \frac{c+a-2b}{a+b} \geq 0$ , where a , b , c are all positive real numbers. I wanted to prove it but had some difficult time , seeing no connection to any known standard inequalities I began to simplify the expression multiplied by $(a+b)(b+c)(c+a)$ , after some simple and elementary but tedious calculations I obtained:-

$$(a+b)(b+c)(c+a) \left(\frac{a+b-2c}{b+c} + \frac{b+c-2a}{c+a} + \frac{c+a-2b}{a+b}\right) \\=a(a-c)^2 + b(b-a)^2 + c(c-b)^2$$ , which is obviously non-negative thus proving the inequality. But I am wondering , is there any other way(s) to prove the inequality? Also, what is the shortest way of deriving the above said identity ?

share|improve this question
    
you can think of $a,b$ and $c$ as lengths a of triangle, may be you will work it out. –  Mohamez Oct 12 '12 at 10:56
    
@ mohamez:- Um, a,b,c are arbitary positive real numbers , so perhaps they can not loosely be assumed to be the sides of a triangle (e.g.:- a,b,c can take the values 2,3,7 ; but a triangle can not) –  Souvik Dey Oct 12 '12 at 12:39
    
@ Souvik Why not?! since the sides of a triangle are also arbitary positive real numbers. for example take $3$ sticks of lenghts respectively $2m$, $3m$ and $7m$ can you join the vertices to make up a triangle? i'm making no condition on the triangle! Whatch this:youtube.com/watch?v=v5ErHe-ls1A isn't the sticks in the video of arbitary lenghts? –  Mohamez Oct 12 '12 at 18:17
add comment

2 Answers

up vote 1 down vote accepted

We can set $a+b=C$ and so on, then minimize

$$ f(A,B,C)=\sum_{cyc}\frac{2C-B}{A}. $$

If we regard $f$ as a function of $B$ and $C$ only, we have that $\frac{\partial f}{\partial B}=0$ implies $(B^2-AC)(2A-C)=0$, and $\frac{\partial f}{\partial C}=0$ implies $(C^2-AB)(2B-A)=0$. So we have four stationary points:

$$ (B/A,C/A)\in\left\{(1,1),(2,4),(2,1/2),(1/2,1/4)\right\}.$$

Without loss of generality we can additionally assume $A\leq B\leq C$, having stationary points for

$$(A,B,C) = (\lambda,\lambda,\lambda)\quad\mbox{or}\quad(\lambda,2\lambda,4\lambda).$$

We can now substitute these values into $f$ to prove the inequality.

share|improve this answer
add comment

To address the first question, let $x = b+c$, $y = c+a$ and $z = a+b$. Then $2c=x+y-z$, so the inequality rewrites as $$ \frac{y-(x+y-z)}{x} + \frac{z-(y+z-x)}{y} + \frac{x-(x+y-z)}{z} \ge 0 $$ or $$\frac{z}{x} + \frac{x}{y} + \frac{y}{z} \ge 3,$$ which follows immediately from the AM-GM inequality.

share|improve this answer
    
You have mistaken in writing down the inequality , what you have wrote is ((a-c)/(b+c)) + ((b-a)/(c+a)) + ((c-b)/(a+b)) ≥ 0 ; this is not what I stated. Using x=b+c , y=c+a , z=a+b , the inequality I posed becomes 2(z/x + y/z + x/y)≥ 3 + y/x + z/y + x/z ; which is much stronger than AM-GM inequality –  Souvik Dey Oct 17 '12 at 9:35
1  
I totally agree with @Souvik Dey - since the inequality has more than 1 stationary points, it cannot follow from an inequality among means, since these inequalities have the unique stationary point $a=b=\ldots$. –  Jack D'Aurizio Oct 23 '12 at 10:28
    
@JackD'Aurizio:- Nice to see that you noticed the fact that v_Enhance's proof is wrong ; though I have accepted your previous answer I still think that the procedure you adopted is not quite elementary as the inequality itself. Can you give it a try of proving the inequality by proving the alternative version that I stated in my last comment , in an elementary way ? –  Souvik Dey Oct 23 '12 at 12:47
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.