[Thgx0312555]

$3(x^2+(1-x-y)^2+(1-y)^2)$

$= (x^2+(1-x-y)^2+(1-y)^2)+(y^2+(1-y-z)^2+(1-z)^2)+(z^2+(1-z-x)^2+(1-x)^2)$

$= x^2+(1-x)^2+y^2+(1-y)^2+z^2+(1-z)^2+(1-x-y)^2+(1-y-z)^2+(1-z-x)^2$

$\ge x^2+(1-x)^2+y^2+(1-y)^2+z^2+(1-z)^2$

$= (2x^2-2x+1)+(2y^2-2y+1)+(2z^2-2z+1)$

ซึ่งสามารถพิสูจน์ได้โดยกำลังสองสมบูรณ์ว่า $2x^2-2x+1,2y^2-2y+1,2z^2-2z+1 \ge \dfrac{1}{4}$

$\therefore 3(x^2+(1-x-y)^2+(1-y)^2) \ge \dfrac{3}{4}$

$x^2+(1-x-y)^2+(1-y)^2\ge \dfrac{1}{4}$

แทน $x=y=z=\dfrac{1}{2}$ จะได้ $x^2+(1-x-y)^2+(1-y)^2= \dfrac{1}{4}$

ดังนั้นค่าต่ำสุดคือ $\dfrac{1}{4}$