每日一题[663]系数配凑

已知$2x^2+2y^2-xy=1$，求$3x^2+4y^2$的最大值．

分析与解　条件即$$4x^2+4y^2-2xy=2,$$引入含参数$\lambda$的均值不等式$$-2xy\leqslant \lambda x^2+\dfrac{1}{\lambda}y^2,\lambda>0,$$以及$$-2xy\geqslant \lambda x^2+\dfrac{1}{\lambda}y^2,\lambda<0,$$于是考虑系数$$\left(4+\lambda\right):\left(4+\dfrac{1}{\lambda }\right)=3:4,$$解得$\lambda =-\dfrac 32$或$\lambda =\dfrac 12$．于是可得$$\dfrac 56\left(3x^2+4y^2\right)\leqslant 2\leqslant \dfrac 32\left(3x^2+4y^2\right),$$从而可得$3x^2+4y^2$的最大值为$\dfrac{12}5$，最小值为$\dfrac 43$．