Use Cauchy-Schwarz's inequality we have:

\(L.H.S=\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{\dfrac{1}{b}+\dfrac{1}{c}}+\dfrac{1}{\dfrac{1}{c}+\dfrac{1}{a}}\)

\(=\dfrac{1}{\dfrac{a+b}{ab}}+\dfrac{1}{\dfrac{b+c}{bc}}+\dfrac{1}{\dfrac{c+a}{ca}}\)

\(=\dfrac{ab}{a+b}+\dfrac{ac}{a+c}+\dfrac{bc}{b+c}\)

\(=a+b+c-\left(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}\right)\)

\(\le a+b+c-\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}\)

\(=a+b+c-\dfrac{a+b+c}{2}=\dfrac{a+b+c}{2}=R.H.S\)

When \(a=b=c\)

:( another way 

We going to prove this inequality : \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

<=> \(\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\)

<=> \(\left(a+b\right)^2\ge4ab\)

<=> \(a^2+2ab+b^2\ge4ab\)

<=> \(a^2-2ab+b^2\ge0\) 

<=> \(\left(a-b\right)^2\ge0\) (it's true)

So that : 

\(\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{\dfrac{1}{b}+\dfrac{1}{c}}+\dfrac{1}{\dfrac{1}{c}+\dfrac{1}{a}}\le\dfrac{1}{\dfrac{4}{a+b}}+\dfrac{1}{\dfrac{4}{b+c}}+\dfrac{1}{\dfrac{4}{c+a}}\)

\(..............\le\dfrac{a+b}{4}+\dfrac{b+c}{4}+\dfrac{c+a}{4}=\dfrac{2\left(a+b+c\right)}{4}=\dfrac{a+b+c}{2}\)

When a = b = c