Ta có :
\(P=a+2b=\left(a-2\right)+2\left(b+3\right)-4\)
\(\Rightarrow P+4=\left(a-2\right)+2\left(b+3\right)\)
\(\Rightarrow\left(P+4\right)^2=\left(\left(a-2\right)+2\left(b+3\right)\right)^2\le\left(1^2+2^2\right)\left(\left(a-2\right)^2+\left(b+3\right)^2\right)\)
\(=25\)
\(\Rightarrow-5\le P+4\le5\)
\(\Rightarrow P\ge-9\)
Dấu " = " xảy ra khi \(\frac{a-2}{1}=\frac{b+3}{2},\left(a-2\right)^2+\left(b+3\right)^2=5\)
\(\Rightarrow a-2=-1,b+3=-2\Rightarrow a=1,b=-5\)