Ta có:\(P=\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\left(đkxđ:a,b\ne0\right)\)
\(P=1+\dfrac{a}{b}+\dfrac{b}{a}+1\)
\(P=2+\dfrac{a}{b}+\dfrac{b}{a}\)
Vì a,b là 2 số cùng dấu\(\Rightarrow\dfrac{a}{b};\dfrac{b}{a}>0\)
AM-GM:
\(\Rightarrow\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}\cdot\dfrac{b}{a}}=2\)
\(\Rightarrow P\ge2+2=4\)
\(\Rightarrow MINP=4\Leftrightarrow a=b\)