Applying Cauchy's inequaility , we have 

\(\dfrac{ab}{a+b}\le\dfrac{\dfrac{\left(a+b\right)^2}{4}}{a+b}=\dfrac{a+b}{4}\)

\(\dfrac{bc}{b+c}\le\dfrac{\dfrac{\left(b+c\right)^2}{4}}{b+c}=\dfrac{b+c}{4}\)

\(\dfrac{ca}{c+a}\le\dfrac{\dfrac{\left(c+a\right)^2}{4}}{c+a}=\dfrac{c+a}{4}\)

So \(\sum\left(\dfrac{ab}{a+b}\right)\le\dfrac{a+b+b+c+c+a}{4}=\dfrac{2\left(a+b+c\right)}{4}=\dfrac{1}{2}\left(a+b+c\right)\)

Equation occur

<=> a = b = c 

Another way :

Applying Cauchy's inequality , we have 

\(\dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ca}{c+a}\le\dfrac{ab}{2\sqrt{ab}}+\dfrac{bc}{2\sqrt{bc}}+\dfrac{ca}{2\sqrt{ca}}=\dfrac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}\)

We have an extra inquality : With a,b,c are positive then \(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)

Prove: \(2\left(a+b+c\right)\ge2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)

<=>  \(\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2+\left(\sqrt{c}-\sqrt{a}\right)^2\ge0\)  (Right) 

So \(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)  is true

We have 

\(\dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ca}{c+a}\le\dfrac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}\le\dfrac{1}{2}.\left(a+b+c\right)\)

Equation occur <=> a = b = c