Extremum of $f (x,y) := x^2+y^2$ subject to the equality constraint $x+y=3$ I had to find the extremum of $z=x^2+y^2$ subject to the constraint $x+y=3$. I used Lagrange multipliers to reach the conclusion that $(1.5,1.5)$ is an extremum point, but had no way of determining whether it's a maximum or a minimum (we did not study the Sylvester criteria). Regardless, intuitively, the most symmetric sum usually gives the largest result, and this is what I used as a justification for the point being a maximum. This is, of course, hardly a mathematical way of showing the correctness of a statement, which is why I ask here what way there is to show it's a maximum in a correct well defined orderly fashion?

By CS inequality:
$$
x+y=(x,y)\cdot (1,1)\le \sqrt{x^2+y^2}\sqrt{2}
$$
Since $x+y=3$:
$$
3\le \sqrt{2}\sqrt{x^2+y^2}
$$
Now, squaring both sided yields
$$
9\le 2(x^2+y^2)
$$
In other words
$$
x^2+y^2\ge \frac{9}{2}
$$
This lower bound is attained when $x=y=3/2$.
