The enquality equivalent : $ |\frac{\sum_{sym}a^{2}.b-6abc}{a^{2}.c+c^{2}.b+b^{2}.a-a^{2}.b-b^{2}.c-c^{2}.a}| > 1$. $ \Leftrightarrow |\sum_{sym}a^{2}.b-6abc| > |a^{2}.c+c^{2}.b+b^{2}.a-a^{2}.b-b^{2}.c-c^{2}.a|$.(1) By AM-GM , $ \sum_{sym}a^{2}.b\ge 6abc$. So (1)$ \Leftrightarrow\sum_{sym}a^{2}.b-6abc>|a^{2}.c+c^{2}.b+b^{2}.a-a^{2}.b-b^{2}.c-c^{2}.a|$.(2) We consider two cases Case 1 $ a^{2}.c+c^{2}.b+b^{2}.a > a^{2}.b+b^{2}.c+c^{2}.a$. Then (2)$ \Leftrightarrow\sum_{sym}a^{2}.b-6abc > a^{2}.c+c^{2}.b+b^{2}.a-a^{2}.b-b^{2}.c-c^{2}.a$ Or $ 2(a^{2}.b+b^{2}.c+c^{2}.a) > 6abc$ It's right ( by AM - GM ), and the equal sign not hold because $ a,b,c$ are diferent numbers . Case 2 $ a^{2}.c+c^{2}.b+b^{2}.a < a^{2}.b+b^{2}.c+c^{2}.a$. Similary. Done

Omid Hatami wrote: $ a,b,c$ are three different positive real numbers. Prove that: \[ \left|\frac{a+b}{a-b}+\frac{b+c}{b-c}+\frac{c+a}{c-a}\right|>1\] Easy... $ \frac{a+b}{a-b}= x;\frac{b+c}{b-c}= y;\frac{a+c}{c-a}= z$ $ \rightarrow xy+yz+xz = 1$ $ (x+y+z)^{2}\geq 3(xy+yz+xz) = 3\rightarrow |x+y+z|\geq\sqrt{3}> 1$

Phạm Thành Quang wrote: Omid Hatami wrote: $ a,b,c$ are three different positive real numbers. Prove that: \[ \left|\frac{a+b}{a-b}+\frac{b+c}{b-c}+\frac{c+a}{c-a}\right|>1\] Easy... $ \frac{a+b}{a-b}= x;\frac{b+c}{b-c}= y;\frac{a+c}{c-a}= z$ $ \rightarrow xy+yz+xz = 1$ $ (x+y+z)^{2}\geq 3(xy+yz+xz) = 3\rightarrow |x+y+z|\geq\sqrt{3}> 1$ Very nice .These problem is an old inequality .

Just some corrections. Actually, $ xy+yz+zx =-1$. But this is not a big deal, since $ (x+y+z)^{2}= x^{2}+y^{2}+z^{2}+2(xy+yz+zx) = x^{2}+y^{2}+z^{2}-2 > 1$, because $ x^{2}$, $ y^{2}$ and $ z^{2}$ are all bigger than $ 1$. Note also that the bound cannot be improved: if $ a = c^{3}$ and $ b = c^{2}$, then $ x = y =\frac{c+1}{c-1}$ and $ z =\frac{1-c^{2}}{1+c^{2}}$. Letting $ c\to\infty$, we have $ x,y\to 1$ and $ z\to-1$, and $ x+y+z$ goes to $ 1$.

There is also another reasoning: Obviously the inequality is equivalent to proving $ \sum_{sym}a^{2}b-6abc\geq |(a-b)(b-c)(c-a)|$ now if we replace each $ a,b,c$ with $ a-t,b-t,c-t$ where $ t\le\min{a,b,c}$ then the LHS decreases while RHS doesnt change thus we can safely suppose $ c=0$ then we are left with $ a+b\geq |a-b|$ which is easy...

there is a (somehow) similar problem: determine the maximum value of: $ f(a,b,c)=\left|\frac {a - b}{a + b} + \frac {b - c}{b + c} + \frac {c - a}{c + a}\right|$ where $ a,b,c$ are the side lengths of a nondegenerate triangle.

owh ..

Very nice inequality.

Attachments:
123.doc (117kb)

This is second my second solution.

Attachments:
solution2.doc (47kb)
solution2.doc (47kb)

Zarif wrote: This is second my second solution. But it's already posted. Can't you read the above posts?

Pain rinnegan wrote: But it's already posted. Can't you read the above posts? It's not like anything bad happens if the same solution is posted twice... Furthermore it can be helpful to ones proof writing skills to write down the proof, and the site is a nice place to save the solutions. @Zarif: $ xy+yz+zx=-1$ in your second solution, so it is wrong. Se cyshine correction further up for more details. It's more convenient to other users if your write it in LaTeX and post it directly on the site. See http://www.mathlinks.ro/LaTeX/AoPS_L_About.php.[/url]

Let $x = \frac{a+b}{a-b}$ , $y =\frac{b+c}{b-c}$ , $z=\frac{c+a}{c-a}$ . It is easy to see $xy+yz+zx=1$ . We can see $(x+y+z)^2 \geq 3(xy+yz+zx) = 3$ . Now we can see $$|x+y+z| \geq \sqrt{3} > 1$$. $\blacksquare$