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Hello I just wondering if it's possible to prove this inequality: there are positive, various $a,b,c$ and

$ \frac{3a-b}{3} \ge x \ge \frac{3(a^2-b^2)}{3a+b}$

$ \frac{3b-c}{3} \ge y \ge \frac{3(b^2-c^2)}{3b+c}$

$ \frac{3c-a}{3} \ge z \ge \frac{3(c^2-a^2)}{3c+a}$

I want to prove that $x+y \ge z$ when we add we get

$\frac{3a+2b-c}{3} \ge x+y \ge \frac {3(a-b)(a+b)}{3 a+b}+\frac {3(b-c)(b+c)}{3 b+c} $ but further I don't know how to compare it with $z$

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1 Answer 1

Hint:

To prove that $x+y\ge z$ we need to prove that minimum value of $x$ plus minimum value of $y$ is $\ge$ maximum value of $z$.

So you have:

$x=\frac{3(a^2-b^2)}{3a+b}; y=\frac{3(b^2-c^2)}{3b+c}; z=\frac{3c-a}{3}$

Just prove that $\frac{3(a^2-b^2)}{3a+b}+\frac{3(b^2-c^2)}{3b+c}\ge\frac{3c-a}{3}$

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ok, thank you :) –  Marco Nov 14 '13 at 16:30
    
let me know how it's going :) –  Gintas K Nov 14 '13 at 16:37
    
@Marco I solved this and got: wolframalpha.com/input/… take a look :) So this equality is correct only if a,b and c meets some conditions :) –  Gintas K Nov 14 '13 at 16:49
    
looks a bit complicated, and only the second condition satisfy the assumptions –  Marco Nov 14 '13 at 16:56
    
So we proved that this equality is incorrect with the given limits of x,y,z :) –  Gintas K Nov 14 '13 at 16:58

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