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 !!!

With Lagrange multipliers: consider $f(x,y,z)=(x^2+1)(y^2+1)(z^2+1)$ and
$$
L(x,y,z,t)=f(x,y,z)-t(x+y+z-3)
$$
Then you have
\begin{align}
\frac{\partial L}{\partial x}&=2x(y^2+1)(z^2+1)-t \\[1ex]
\frac{\partial L}{\partial x}&=2y(x^2+1)(z^2+1)-t \\[1ex]
\frac{\partial L}{\partial x}&=2z(x^2+1)(y^2+1)-t
\end{align}
These should vanish together, so we obtain, for instance,
$$
2x(y^2+1)(z^2+1)=2y(x^2+1)(z^2+1)
$$
that yields $xy^2+x=x^2y+y$, hence $xy(x-y)=x-y$. Thus either $xy=1$ or $x=y$. Similarly, $xz=1$ or $x=z$. Thus we have four cases to examine.
The case $x=y$ and $y=z$ yields $x=y=z=1$.
The case $xy=1$ and $xz=1$ yields $y=z$ and so $x=2,y=1/2,z=1/2$.
Since the function $f$ as well as the constraint are symmetric in $x,y,z$, the remaining two cases will provide $x=1/2,y=2,z=1/2$ and $x=1/2,y=1/2,z=2$.
We have $f(1,1,1)=8$ and $f(2,1/2,1/2)=125/16$.
