Áp dụng BĐT Bunyakovsky ta có:
P^{2}\leq (x^{2}+y^{2})[y(2x+3y)+x(3x+2y)]\leq 2.(4xy+3x^{2}+3y^{2})\leq 6(x^{2}+y^{2})+8xy\leq 6.2+4.2=20
P\leq \sqrt{20}
Dấu "=" xảy ra \Leftrightarrow \left\{\begin{matrix} \frac{x}{\sqrt{y(2x+3y)}}=\frac{y}{\sqrt{x(3x+2y)}}\\ x^{2}+y^{2}=2 \end{matrix}\right...