My mathematics textbook lists the following example:
If $\ a,b,c\in ℝ\ $ and $\ a^2+b^2+c^2=1\ $ then the minimum and maximum value of $\ ab+bc+ca\ $ is:
(A) $\ \frac{1}{2}, 1\ $
(B) $\ 1, 4\ $
(C) $\ \frac{1}{2}, \frac{3}{2}\ $
(D) $\ -\frac{1}{2}, 1\ $
To which, it says:
Given, $\ a^2+b^2+c^2=1$
We know that $\ (a+b+c)^2\geq0\ $
$\ \implies a^2+b^2+c^2+2ab+2bc+2ca\geq0$
$\ \implies 1+2(ab+bc+ca)\geq0$
$\ \implies ab+bc+ca\geq\frac{-1}{2}$
Also, $\ (b-c)^2+(c-a)^2+(a-b)^2\geq0\ $
$\ \implies 2(a^2+b^2+c^2)-2(ab+bc+ca)\geq0$
$\ \implies 2(1)\geq2(ab+bc+ca)$
$\ \implies ab+bc+ca\leq1$
$\therefore\ $(D) $\ -\frac{1}{2}, 1\ $ is the correct answer.
However, I fail to understand how that puts us in the confidence that no other interval exists for $\ ab+bc+ca\ $. Simply put, just like simplifying $\ (a+b+c)^2\ $ and re-factoring $\ (b-c)^2+(c-a)^2+(a-b)^2\ $ yield two different inequality expressions, how do we know that factoring some other way won't result in an interval that might affect the maximum or minimum possible value?
Set $f(x,y,z)=xy+yz+zx.$ We know already that $$-\frac{1}{2}\le f(x,y,z) \le 1.$$
It suffices to find $(x,y,z)$ such that the values $-\frac{1}{2}$ and $1$ are obtained.
If $x=y=z,$ then the given constraint rewrites $$1=x^2+y^2+z^2=3x^2.$$ Also, $$f(x,x,x)=3x^2=1.$$ This maximal value is obtained in points $$\left(-\frac{1}{\sqrt3},-\frac{1}{\sqrt3},-\frac{1}{\sqrt3}\right) \quad\text{and} \quad \left(\frac{1}{\sqrt3},\frac{1}{\sqrt3},\frac{1}{\sqrt3}\right).$$
For the minimum, choose one variable nul and two others opposite (we want to obtain negative product), WLOG $$(x,y,z)=(x,-x,0).$$ Then $$1=x^2+y^2+z^2=2x^2$$ and $$f(x,-x,0)=-x^2=-\frac{1}{2}.$$