Cauchy Schwarz - Finding minimum of a function For $x, y , z$ in real numbers, If $2x+y+z=5$, then what is the min value of $x^2 + y^2 + z^2 - 2x + 4y + 6$.  This is a weekend brain teaser in the 2nd week of Calc 3.

This problem can be solved by using elementary math only (no derivatives, no Lagrange multipliers, no C-S ineqaulity etc). 
We have to minimize the expression:
$$f=x^2+y^2+z^2-2x+4y+6=(x-1)^2+(y+2)^2+z^2+1$$
Introduce substitution:
$$u=x-1, \ \ v=y+2, \ \ w=z$$
The expression that has to be minimized now becomes:
$$f=u^2+v^2+w^2+1\tag{1}$$
...with the following constraint:
$$2u+v+w=5\tag{2}$$
Make one more subsitution: 
$$v=x-y, \ \ w=x+y$$
Replace that into (1) and (2) and you get:
$$f=u^2+2x^2+2y^2+1\tag{3}$$
...and the constraint becomes:
$$2u+2x=5\tag{4}$$
The point is: contraint (4) give us the possibility to minimize (3) by simply choosing $y=0$. Now the problem is much simpler. We have to minimize:
$$f=u^2+2x^2+1\tag{5}$$
...with constraint:
$$u+x=\frac52\tag{6}$$
From (6):
$$x=\frac 52-u$$
...so (7) becomes:
$$f=u^2+2(\frac 52 - u)^2+1=3(u-\frac53)^2+\frac{31}{6}$$
So the minimum value is $f_{min}=\frac{31}{6}$ and it is reached for $u=\frac53$.
