If $a^2+b^2+c^2=1$ then find the range of $ab+bc+ca$.
My attempt: We know $$ 0\le(a + b+c)^2 $$ $$ 0\le a^2+b^2+c^2+2(ab+bc+ca) $$ $$ 0\le1 + 2(ab+bc+ca) $$ $$ -1/2\le ab+bc+ca $$ $$ range = [-1/2 ,\infty) $$ But the answer is $[-1/2 , 1]$. I would like to know how to proceed further.
On
By Cauchy-Schwartz, $$ab+bc+ca = \langle(a,b,c), (b,c,a)\rangle \leq \sqrt{a^2+b^2+c^2}\sqrt{b^2+c^2+a^2} = a^2+b^2+c^2 = 1.$$
On
Note that $$ a^2+b^2+c^2+2(ab+bc+ca) = (a + b + c)^2 $$ (which is not $(a^2 + b^2+c^2)^2$).
The points such that $a^2 + b^2+c^2 = 1$ all lie on a sphere of radius $1.$ You can never be more than $1$ unit away from the origin on such a sphere, so $a \leq 1,$ $b \leq 1,$ and $c \leq 1.$
These three facts already mean that $a + b + c \leq 3,$ and therefore $(a + b + c)^2 \leq 9.$ So you have a (not very large!) upper bound as well as a lower bound.
The question is exactly what the least upper bound is. Looking at a sphere, it should be clear that the sum $a + b + c$ cannot actually get close to $3.$ You can use various techniques to try to find the maximum of $a + b + c$; one way is to recognize that it will occur at the point where the sphere $a^2 + b^2+c^2 = 1$ is tangent to a plane with equation $a + b + c = k.$ In order for $(a,b,c)$ to be the point of tangency, the vector from the origin to $(a,b,c)$ must be normal to the plane $a + b + c = k,$ and therefore $a = b = c.$ But then $3a^2 = a^2 + b^2+c^2 = 1,$ so $a = \sqrt{\frac13},$ and therefore $b = c = \sqrt{\frac13}$ as well. From this you can find the maximum value of $(a + b + c)^2.$
So that's an answer based on geometric intuition. But note that Cauchy-Schwartz (used in other answers) gets you this result a lot quicker.
Use that $$x^2+y^2+z^2\geq xy+yz+zx$$ for all real numbers $$x,y,z$$