Find the minimum and maximum value of $x+y+z+xy+yz+xz$ if
$x^2+y^2+z^2 = 1$
I converted it to uvw,
$3u+3v^2$ is the expression, $(3u)^2-2(3v^2)=1$, is the constraint.
now I don't know what to do, I'm still learning the uvw method and I don't know if we can use the basic theorem to solve this or Tej's theorem.
On
Let $x+y+z=3u$, $xy+xz+yz=3v^2,$ where $v^2$ can be negative, and $xyz=w^3$.
Thus, the condition gives $9u^2-6v^2=1$, which does not depend on $w^3$.
Also, the expression $x+y+z+xy+xz+yz=3u+3v^2$ does not depend on $w^3$, which says that
it's enough to find an extreme value of our expression for the extreme value of $w^3$,
which happens for equality case of two variables.
Let $y=x$.
Thus, $z^2=1-2x^2$ and we need to find an extreme value of $$2x+x^2+z(1+2x),$$ where $z^2=1-2x^2$ and $-\frac{1}{\sqrt2}\leq x\leq \frac{1}{\sqrt2}.$
I got $$-1\leq x+y+z+xy+xz+yz\leq1+\sqrt3.$$ By the way, I think using $uvw$ for this problem is not so necessary.
Here there are much more simpler ways. For example, see the Batominovski's comment.
$$3u+3v^2=3u+\dfrac{(3u)^2-1}2=\dfrac{9u^2+6u-1}2=\dfrac{(3u+1)^2-2}2$$
Now for real $u,(3u+1)^2\ge0$