Tìm Min $M=\sqrt{x^2+2y^2-6x+4y+11}+\sqrt{x^2+3y^2+2x+6y+4}$

Bài toán:

Tìm Min $M=\sqrt{x^2+2y^2-6x+4y+11}+\sqrt{x^2+3y^2+2x+6y+4}$

Lời giải:

$+)x^2+2y^2-6x+4y+11=(x-3)^2+2(y+1)^2\\ +)x^2+3y^2+2x+6y+4=(x+1)^2+3(y+1)^2$

Có:

$M=\sqrt{(x-3)^2+2(y+1)^2}+\sqrt{(x+1)^2+3(y+1)^2}$

$\geq \sqrt{(x-3)^2}+\sqrt{(x+1)^2}\\ =|x-3|+|x+1|$

$=|3-x|+|x-1|\ge |3-x+x+1|=4$

Dấu bằng: $y=-1;-1\le x\le 3$

Vậy $minM=4$ khi $y=-1;-1\le x\le 3$