# Minimize $\;\left(-x+y+1 \right)^2 + \left( x-y-2\right)^2 + \left(x+2y-3 \right)^2 \;$ without using partial derivatives

How to find minimum of the expression $$\, \big(\!-x+y+1 \big)^2 + \big( x-y-2\big)^2 + \big(x+2y-3 \big)^2 \,$$ without using partial derivatives?

It is easy to find the answer $\; x = 2, \; y = \dfrac{1}{2}\;$ by computing gradient of the expression above, but I do not see the way to do so without using partial derivatives.

This is the part of homework for linear algebra class designed for freshmen. I feel that there is no way firs-year students are expected to use partial derivatives because this topic is only taught in the end of CALC II class. I feel very dumb and discouraged since I could not help the student. We tried to make substitution $\;z = x-y,\;$ or to expand the brackets, but nothing seemed to give definite answer. Any help is appreciated.

By completing the square, $$\big(-x+y+1 \big)^2 + \big( x-y-2\big)^2 + \big(x+2y-3 \big)^2 = 3(x-2)^2+6(y-\frac{1}{2})^2 +\frac{1}{2}$$ the minimun is $\frac{1}{2}$
$a = y-x+1, b = x-y-2, c = x+2y-3$, then find an equation in $a,b,c$ and use CS inequality.
since $$f=(-x+y+1)^2+(x-y-2)^2+(x+2y-3)^2\ge (-x+y+1)^2+(x-y-2)^2$$ and Use Cauchy-Schwarz inequality we have $$[(-x+y+1)^2+(x-y-2)^2][1^2+1^2]\ge (-x+y+1+x-y-2)^2=1$$ so $$f\ge \dfrac{1}{2}$$
If you expand the expression you see there are no cross terms, so it boils down to separately minimizing a function of $x$ and a function of $y$.