O outsos Apr 2010 456 1 Oct 20, 2019 #1 prove that: \(\displaystyle (a+b+c)\geq\frac{(\sqrt a+\sqrt b+\sqrt c)^2}{3}\) I tried by expanding the square Last edited: Oct 20, 2019
prove that: \(\displaystyle (a+b+c)\geq\frac{(\sqrt a+\sqrt b+\sqrt c)^2}{3}\) I tried by expanding the square
tahirimanov19 Mar 2015 182 68 Universe 2.71828i3.14159 Oct 20, 2019 #2 outsos said: prove that: \(\displaystyle (a+b+c)\geq\frac{(\sqrt a+\sqrt b+\sqrt c)^2}{3}\) I tried by expanding the square Click to expand... By expanding we get $2a+2b+2c \ge 2 \sqrt{ab} + 2 \sqrt{bc} + 2 \sqrt{ca} \Rightarrow a+b+c \ge \sqrt{ab} + \sqrt{bc} + \sqrt{ca}$. $a+b+c = \dfrac{a+b}{2} + \dfrac{b+c}{2} + \dfrac{c+a}{2}$. Now prove, $\dfrac{a+b}{2} \ge \sqrt{ab}$ Reactions: 3 people
outsos said: prove that: \(\displaystyle (a+b+c)\geq\frac{(\sqrt a+\sqrt b+\sqrt c)^2}{3}\) I tried by expanding the square Click to expand... By expanding we get $2a+2b+2c \ge 2 \sqrt{ab} + 2 \sqrt{bc} + 2 \sqrt{ca} \Rightarrow a+b+c \ge \sqrt{ab} + \sqrt{bc} + \sqrt{ca}$. $a+b+c = \dfrac{a+b}{2} + \dfrac{b+c}{2} + \dfrac{c+a}{2}$. Now prove, $\dfrac{a+b}{2} \ge \sqrt{ab}$