Áp dụng BĐT $\color{black}{a^2+b^2≥2aba^2+b2≥2ab}$
$\color{black}{⇒x^2+1≥2x⇒2x^2≥4x−2⇒x^2+1≥2x⇒2x2≥4x−2}$
$\color{black}{y^2+1≥2y⇒3y^2≥6y−3y^2+1≥2y⇒3y^2≥6y−3}$
$\color{black}{⇒2x2+3y^2≥2(2x+3y)−5⇒2x2+3y^2≥2(2x+3y)−5}$
Mà $\color{black}{2x2+3y^2≤52x2+3y^2≤5}$
$\color{black}{⇒2(2x+3y)−5≤5⇒2x+3y≤5⇒2(2x+3y)−5≤5⇒2x+3y≤5}$
Vậy Max $\color{black}{A = 5}$ khi $\color{black}{x = y (= 1)}$
