\(\dfrac{a}{2}+\dfrac{b}{3}=\dfrac{a+b}{2+3}\)(1)

\(\Leftrightarrow\dfrac{3a+2b}{6}=\dfrac{a+b}{5}\)

\(\Rightarrow15a+10b=6a+6b\)

\(\Rightarrow9a+4b=0\)

\(\Rightarrow a=-\dfrac{4b}{9}\)(2)

(1),(2) => \(-\dfrac{4b}{18}+\dfrac{b}{3}=\dfrac{-\dfrac{4b}{9}+b}{2+3}\)

Solve this expression we have b = 0.

So from here we have : \(\dfrac{a}{2}+0=\dfrac{a+0}{5}\Rightarrow a=0\)

So (a;b) = (0;0)

\(\dfrac{a}{2}+\dfrac{b}{3}=\dfrac{a+b}{2+3}=\dfrac{a+b}{5}=\dfrac{a}{5}+\dfrac{b}{5}\)

We have:

\(\dfrac{a}{2}\ge\dfrac{a}{5}\)  (Equal sign occurs  \(\Leftrightarrow a=0\))

\(\dfrac{a}{3}\ge\dfrac{a}{5}\)  (Equal sign occurs  \(\Leftrightarrow a=0\))

\(\Leftrightarrow\dfrac{a}{2}+\dfrac{a}{3}\ge\dfrac{a+b}{5}\) 

Equal sign occurs  \(\Leftrightarrow a=b=0\)