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I have tried the arithmetic-geometric inequality on $(-a+b+c)(a-b+c)(a+b-c)$ which gives

$$(-a+b+c)(a-b+c)(a+b-c) \leq \left(\frac{a+b+c}{3}\right)^3$$

and on $abc$ which gives

$$abc \leq \left(\frac{a+b+c}{3}\right)^3.$$

Since both inequalities have the same righthand side, I have tried to deduce something about the lefthand sides, but to no avail. Can somebody help me, please? I am sure it is something simple I have missed.

  • 2
    In general, if $x < z$ and $y < z$ there is no reason to believe that $x < y$ or $y < x$.2012-07-14
  • 3
    think of $a, b$ and $c$ as lengths of a triangle.2012-07-14
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    @mohamez: below, I derive a simple formula for the ratio of the two sides of this inequality which involves only the circumradius and the distance between the incenter and circumcenter.2012-07-15
  • 0
    You cannot apply AM-GM on $-a+b+c, a-b+c, a+b-c$, unless you assume or prove that $a,b,c$ are the sides of a triangle (so that the terms are all nonnegative).2015-06-03

7 Answers 7