Minimum value of $A = (x^2 + 1)(y^2 +1)(z^2 +1)$? Given: $\begin{cases}x;y;z \in\Bbb R\\x+y+z=3\end{cases}$
Then what is the minimum value of $A = (x^2 + 1)(y^2 +1)(z^2 +1)$ ?
I start with $AM-GM$
So $A \geq 8xyz$
Then I need to prove $xyz\geq 1$ right?
How to do that and if it's a wrong way, please help me to solve it !!!

Let $A(x,y,z)=(1+x^2)(1+y^2)(1+z^2)$, defined on $\{(x,y,z)\in\mathbb{R}^3{\,\mid\,}x+y+z=3\}$.

Since $\max(|x|,|y|,|z|)\ge 3$ would yield $A > 8$, it follows $A$ has a global minimum value, $a$ say.

Since $(x,y,z)=1$ yields $A=8$, it follows that $a\le 8$.

Suppose $A(x,y,z)=a$.

Without loss of generality, assume $x\le y\le z$.

From $x+y+z=3$, it follows that $z\ge 1$.

We must have $\max(|x|,|y|,|z|)< 3$, hence $z < 3$.

Let $s=x+y$.

From $x+y+z=3$ and $1\le z < 3$, we get $0 < s \le 2$.

Let $u=(1+x^2)(1+y^2)$ and let $v=\left(1+\left({\Large{\frac{s}{2}}}\right)^2\right)^2$.

Replacing $y$ by $s-x$, we get
$$
u-v
=
\Bigl({\small{\frac{1}{16}}}\Bigr)
\!
\left((s-2x)^2+2(4-s^2)\right)
(s-2x)^2
$$
hence $u\ge v$ with equality if and only if $s=2x$.

But we can't have $u > v$, else $A(x,y,z) > A\Bigl({\Large{\frac{s}{2}}},{\Large{\frac{s}{2}}},z\Bigr)$, contrary to $A(x,y,z)=a$.

Thus $u=v$, hence $s=2x$, so $y=x$.

It follows that $a$ is the minimum value of 
$$
f(x)
=
A(x,x,3-2x)
=
(1+x^2)^2\left(1+(3-2x)^2\right)
$$
where $f$ can be restricted to the interval $0\le x\le 1$.

At the endpoints we have $f(0)=10$ and $f(1)=8$.

Identically we have
$$
f'(x)
=
12(2x-1)(1+x^2)(x-1)^2
$$
which on the interval $(0,1)$ has the unique critical point $x={\large{\frac{1}{2}}}$.

Noting that $f\Bigl({\large{\frac{1}{2}}}\Bigr)={\large{\frac{125}{16}}}=7.8125 <  8$, it follows that $a={\large{\frac{125}{16}}}$, with $(x,y,z)=\Bigl({\large{\frac{1}{2}}},{\large{\frac{1}{2}}},2\Bigr)$.
