<s>[INST] Find the minimum value of $x_1^2+x_2^2+x_3^2+x_4^2$ subject to $x_1+x_2+x_3+x_4=a$ and $x_1-x_2+x_3-x_4=b$. Question: Find the minimum value of $x_1^2+x_2^2+x_3^2+x_4^2$ subject to $x_1+x_2+x_3+x_4=a$ and $x_1-x_2+x_3-x_4=b$.
My attempt: It can be easily seen that $x_1+x_3=\frac{a+b}{2}$ and $x_2+x_4=\frac{a-b}{2}$. Further, the expression $[x_1^2+x_2^2+x_3^2+x_4^2]$ can be written as $[(x_1+x_3)^2+(x_2+x_4)^2-2(x_1x_3+x_2x_4)].$ I'm having trouble eliminating $(x_1x_3+x_2x_4)$ from this expression. Failing to make any sense out of this, I manipulated the existing expressions to deduce $$x_1x_2+x_1x_4+x_2x_3+x_3x_4=\frac{a^2-b^2}{4}$$and $$(x_1^2+x_3^2)-(x_2^2+x_4^2)+2(x_1x_3-x_2x_4)=a\cdot b$$Beyond this, I cannot make sense of the expressions anymore. I have no idea how to proceed with simplifying the expressions further, and would appreciate hints in the same direction.
 [/INST] By your work and by C-S
$$x_1^2+x_2^2+x_3^2+x_4^2\geq\frac{1}{2}\left(\frac{a+b}{2}\right)^2+\frac{1}{2}\left(\frac{a-b}{2}\right)^2=\frac{a^2+b^2}{4}.$$
The equality occurs for $x_1=x_3=\frac{a+b}{4}$ and $x_2=x_4=\frac{a-b}{4},$ which says that we got a minimal value.
We used the following C-S:
$$x^2+y^2=\frac{1}{2}(1^2+1^2)(x^2+y^2)\geq\frac{1}{2}(x+y)^2.$$
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