# Algebraic Inequality involving AM-GM-HM

If $$a,b,c \;\epsilon \;R^+$$ Then show, $$\frac{bc}{b+c} + \frac{ab}{a+b} + \frac{ac}{a+c} \leq \frac{1}{2} \Bigl(a+b+c\Bigl)$$ I have solved a couple of problems using AM-GM but I have always struggled with selecting terms for the inequality.
I tried writing $a+b+c$ as $x$ and re-writing the inequality as $$\frac{bc}{x-a} + \frac{ab}{x-c} + \frac{ac}{x-b} \leq \frac{1}{2} \Bigl(x\Bigl)$$ This didn't help so instead I wrote it as $$2\Biggl(\frac {1}{a} + \frac{1}{b} + \frac{1}{c}\Biggl ) \leq \frac{1}{2} \Bigl(a+b+c\Bigl)$$ Not sure what to to do next. Any help would be appreciated.

Hint: Remember that $HM(a, b)$ can also be written as $\dfrac {2ab}{a+b}$, which looks a lot like the LHS...

• Oh, alright thank you so much. – Prakhar Nagpal Jul 24 '18 at 7:57
• @PrakharNagpal You're welcome! – Ovi Jul 24 '18 at 7:57

# Proof

Just by $H_n \leq A_n$, we have

\begin{align*} \frac{bc}{b+c} + \frac{ab}{a+b} + \frac{ac}{a+c}&=\frac{1}{2}\left(\frac{2}{\frac{1}{b}+\frac{1}{c}}+\frac{2}{\frac{1}{a}+\frac{1}{b}}+\frac{2}{\frac{1}{a}+\frac{1}{c}}\right)\\&\leq \frac{1}{2}\left(\frac{b+c}{2}+\frac{a+b}{2}+\frac{a+c}{2}\right)\\&=\frac{a+b+c}{2}. \end{align*}

Multiply both sides by $4$ and rearrange: $$\frac{bc}{b+c} + \frac{ab}{a+b} + \frac{ac}{a+c} \leq \frac{1}{2} \Bigl(a+b+c\Bigl) \Rightarrow \\ \left(a+b-\frac{4ab}{a+b}\right)+\left(b+c-\frac{4bc}{b+c}\right)+\left(c+a-\frac{4ca}{c+a}\right)\ge 0 \Rightarrow \\ \frac{(a-b)^2}{a+b}+\frac{(b-c)^2}{b+c}+\frac{(c-a)^2}{c+a}\ge 0.$$

• +1 Nice, refreshing way to do it – Ovi Jul 24 '18 at 17:55
• @Ovi, thank you – farruhota Jul 24 '18 at 18:03

By C-S $$\sum_{cyc}\frac{bc}{b+c}\leq\sum_{cyc}\frac{bc}{(1+1)^2}\left(\frac{1^2}{b}+\frac{1^2}{c}\right)=\frac{a+b+c}{2.}$$