Let $a\ge 0, b\ge 0$ and $c\le 0$ We consider the polynomial $P(x)=ax_1+bx_2^2+cx_3^2$ with variables $x=(x_1,x_2,x_3)\in \mathbb{R}^3$ Is the following equivalence true?
$P$ does not have any local minimum if and only if $a-c\not=0$
Let $a\ge 0, b\ge 0$ and $c\le 0$ We consider the polynomial $P(x)=ax_1+bx_2^2+cx_3^2$ with variables $x=(x_1,x_2,x_3)\in \mathbb{R}^3$ Is the following equivalence true?
$P$ does not have any local minimum if and only if $a-c\not=0$
On
A local minimum has to happen at a point where the gradient $\nabla P=\vec{0}$.
Here $\nabla P=\left<a,2bx_2,2cx_3\right>$.
you can immediately conclude that $a=0$ so $P=bx_2^2+cx_3^2$ which iseither a quadric surface when projected above the $x_2x_3$-plane or degenerate with $P=bx_2^2$ or $P=cx_3^2$. Which gives 3 cases that have local minimums:
It is very easy to determine where $P$ has a local minimum, since the contributions of each variable are separate. If $a\neq 0$, then $P(x)$ can be increased by increasing $x_1$, so $P$ has no local minimum unless $a=0$. Similarly, if $c\neq 0$, then $P(x)$ can always be increased by increasing the absolute value of $x_3$, so $P$ has no local minimum unless $c=0$. Finally, if $a=c=0$, we have $P(x)=bx_2^2$, which is minimized everywhere if $b=0$ and is minimized when $x_2=0$ if $b>0$.
So, $P$ has a local minimum iff $a=c=0$, which under your assumptions is equivalent to $a-c=0$.