How to prove that a Local minimum is Absolute minimum in $R^3$ Just trying to solve this question:
$f(x,y,z) = x^2 + y^2 +3z^2 -xy +2xz+ yz$.
Found the only critical point of the function and explain why she is an absolute minimum.
We learn at class how to found that a point is a critical point so i found her and its $(0,0,0)$.
I know how to say that its a Local minimum, but i don't know how i am supposed to explain why its an absolute minimum and i want to know how to approach it.
 A: I don't know if there's some immediate way to know that the local minimum is a global minimum, and I would like to see such answers if there are.
I would do some algebra to rewrite $f$ as:
$f(x,y,z) = 0.5(x-y)^2 + 0.5(y+z)^2 +(x+z)^2-0.5x^2+1.5z^2$
Then, the only potentially problematic term is $-0.5x^2$.
But minimizing $(x+z)^2-0.5x^2+1.5z^2$ over $z$, which occurs when $2(x+z)+3z=0$, i.e. $z=-0.4x$, we see that:
$(x+z)^2-0.5x^2+1.5z^2 \ge 0.36x^2 -0.5x^2 +0.24x^2 = 0.1x^2 \ge 0$
Therefore $f(x,y,z) \ge 0$
Note that, when I minimized $(x+z)^2-0.5x^2+1.5z^2$ over $z$ to show that $(x+z)^2-0.5x^2+1.5z^2 \ge  0.1x^2$, that was only to derive that inequality. I was not trying to minimize $f$ over only $z$, while ignoring $y$.
Another way to derive the same inequality would be to let $t = z + 0.4x$, with the choice inspired by the fact that we derived above that $z = -0.4x$ minimizes that expression.
Then we have $z = t - 0.4x$ and $z^2 = t^2 - 0.8tx + 0.16x^2$, and:
\begin{align}
(x+z)^2-0.5x^2+1.5z^2 &= x^2 + 2xz + z^2 - 0.5x^2+1.5z^2 \\
& = 0.5x^2 +2xz+2.5z^2 \\
& = 0.5x^2 +2x(t - 0.4x)+2.5(t^2 - 0.8tx + 0.16x^2) \\
& = 0.1x^2 +2.5t^2 \\
&\ge 0.1x^2
\end{align}
Using this inequality in $f$ gives:
\begin{align}
f(x,y,z) &= 0.5(x-y)^2 + 0.5(y+z)^2 +(x+z)^2-0.5x^2+1.5z^2 \\
&\ge 0.5(x-y)^2 + 0.5(y+z)^2 + 0.1x^2 \\
&\ge 0 \qquad \text{ since it is the sum of squares}\\
\end{align}
