Let $a,b,$ and $c$ be positive real numbers, prove that $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \geq 2 \left (\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a} \right) \geq \dfrac{9}{a+b+c}$.
Should I use AM-GM for the expression in the middle of the inequality? We have $a+b \geq 2\sqrt{ab}$ etc.?
Best Answer
Use AM-HM Inequality for both.
$$\frac{a+b}{2}\ge \frac{2}{\frac{1}{a}+\frac{1}{b}} \Rightarrow \frac{1}{a}+\frac{1}{b}\ge \frac{4}{a+b}$$
Similarly, you get $$\frac{1}{a}+\frac{1}{c}\ge \frac{4}{a+c}$$ and $$\frac{1}{c}+\frac{1}{b}\ge \frac{4}{c+b}$$
Now add the three and get the left inequality.
For right inequality, $$\frac{\frac{a+b}{2}+\frac{b+c}{2}+\frac{c+a}{2}}{3}\ge \frac{3}{\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}}$$ or $$\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}\ge \frac{9}{a+b+c}$$