GTLN:
Áp dụng BĐT \(a^2+b^2\ge2ab\)
\(\Rightarrow x^2+1\ge2x\Rightarrow2x^2\ge4x-2\)
\(y^2+1\ge2y\Rightarrow3y^2\ge6y-3\)
\(\Rightarrow2x^2+3y^2\ge2\left(2x+3y\right)-5\)
mà \(2x^2+3y^2\le5\)
\(\Rightarrow2\left(2x+3y\right)-5\le5\Rightarrow2x+3y\le5\)
Vậy Max A = 5 khi x = y = 1