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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.

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    $\begingroup$ Is $$x,y,z$$ assumed to be positive? $\endgroup$ – Dr. Sonnhard Graubner Sep 11 '18 at 13:11
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    $\begingroup$ The maximum can be found via the inequality $$a^2+b^2+c^2\geq bc+ca+ab\,.$$ That is, $$x+y+z\leq \sqrt{3\,\left(x^2+y^2+z^2\right)}=\sqrt{3}$$ and $$yz+zx+xy\leq x^2+y^2+z^2=1\,,$$ so $$x+y+z+yz+zx+xy\leq \sqrt{3}+1\,,$$ where the equality occurs if and only if $x=y=z=\dfrac{1}{\sqrt{3}}$. $\endgroup$ – Batominovski Sep 11 '18 at 13:24
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$$3u+3v^2=3u+\dfrac{(3u)^2-1}2=\dfrac{9u^2+6u-1}2=\dfrac{(3u+1)^2-2}2$$

Now for real $u,(3u+1)^2\ge0$

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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.

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