We have to prove \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

Let it be : \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)(1).

We have (1) <=> \(\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\Leftrightarrow\left(a+b\right)^2\ge4ab\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\), true with all a,b > 0.

Done ! 

We have:

\(\left(a-b\right)^2\ge0\)

=> a² + b² \(\ge2ab\)

=> a² + b² + 2ab \(\ge2\left(2ab\right)=4ab\)

=> \(\left(a+b\right)^2\ge4ab\)

Because a > 0, b > 0 => a+b > 0

\(\Rightarrow\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\Leftrightarrow\dfrac{a}{ab}+\dfrac{b}{ab}\ge\dfrac{4}{a+b}\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

So, .......