Question

x，y，z大于0。x+y+z等于1。求x平方+3y平方+z平方最小值

Answer 1

这是一道经典的数学问题，可以通过代数的方法来求解。

设x+y+z=1x+y+z=1，则有：

\begin{aligned} x &= 1-y \\ y &= 1-z \end{aligned}
x
y
​

=1−y
=1−z
​

将上式代入x^2+3y^2+z^2x
2
+3y
2
+z
2
中，可以得到：

\begin{aligned} x^2+3y^2+z^2 &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y)^2 \\ &= x^2+3y^2+(1-y