An inequality concerning three positive numbers If $a, b, c$ belong to positive real numbers then prove that 
$$\frac{2}{b+c} +\frac{2}{c+a}+\frac{2}{a+b} \le \frac 1a +\frac 1b +\frac 1c.$$
I applied AM-GM inequalities on both sides but I cant even figure out what to do on left side.

Applying the inequality between the Arithmetic Mean and the Harmonic Mean  we get:$$\frac{2}{(a+b)}\le\frac{\frac{1}{a}+\frac{1}{b}}{2}=\frac{a+b}{2ab}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\tag{1}$$
$$\frac{2}{(b+c)}\le\frac{\frac{1}{b}+\frac{1}{c}}{2}=\frac{b+c}{2bc}=\frac{1}{2}\left(\frac{1}{b}+\frac{1}{c}\right)\tag{2}$$
$$\frac{2}{(a+c)}\le\frac{\frac{1}{a}+\frac{1}{c}}{2}=\frac{a+c}{2ac}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{c}\right)\tag{3}$$
On adding (1), (2), and (3) we obtain the required inequality:
$$\frac{2}{(a+b)} +\frac{2}{(b+c)}+\frac{2}{(c+a)} \le \frac{1}{a} +\frac{1}{b }+\frac{1}{c}$$
