Let $ a,b,c$ be positive reals; show that $ \displaystyle a + b + c \leq \frac {bc}{b + c} + \frac {ca}{c + a} + \frac {ab}{a + b} + \frac {1}{2}\left(\frac {bc}{a} + \frac {ca}{b} + \frac {ab}{c}\right)$ (Darij Grinberg)
Problem
Source: Oliforum contest 2009, round 2/4
Tags: inequalities, inequalities proposed
19.10.2009 02:32
19.10.2009 08:06
bboypa wrote: Let $ a,b,c$ be positive reals; show that $ \displaystyle a + b + c \leq \frac {bc}{b + c} + \frac {ca}{c + a} + \frac {ab}{a + b} + \frac {1}{2}\left(\frac {bc}{a} + \frac {ca}{b} + \frac {ab}{c}\right)$ (Darij Grinberg) Let $ a=yz,b=zx$ and $ c=xy,$ where $ x,y,z$ are some positive real numbers. The inequality becomes $ \frac{1}{2}(x^2+y^2+z^2) +xyz \left( \frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right) \ge xy+yz+zx.$ By the Cauchy-Schwarz Inequality, we have $ \frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y} \ge \frac{9}{2(x+y+z)},$ and hence, it suffices to prove that $ x^2+y^2+z^2+\frac{9xyz}{x+y+z} \ge 2(xy+yz+zx),$ which is Schur's Inequality.
19.10.2009 20:16
$ \displaystyle \sum_{cyc}\left(\frac{bc}{b+c}+\frac{bc}{2a}-\frac{b+c}{2}\right)\geq 0$ iff $ \displaystyle \sum_{cyc}\left(\frac{\left(c-a\right)b^2}{2a\left(b+c\right)}-\frac{\left(a-b\right)b^2}{2a\left(b+c\right)}\right)\geq0$ iff $ \displaystyle \sum_{cyc} \left(b-c\right)a^2\left(\frac{1}{c\left(a+b\right)}-\frac{1}{b\left(c+a\right)}\right)\geq0$, which is a sum with three nonnegative addends.