I am trying to show that the derivative of the scalar function $V(\mathbf{x})$, $V'(\mathbf{x})$, is such that
$V'(\mathbf{x})=-(x_1^2+x_2^2)+x_2u\leq-||\mathbf{x}||^2+||\mathbf{x}|| \ |u|$
where $\mathbf{x}=(x_1,x_2)$ and $||\mathbf{x}||=\sqrt{x_1^2+x_2^2}$. My question is: how do we know that $x_2u$ is bounded by $||\mathbf{x}|| \ |u|$?
Observe the following:
$x_2^2 \leq x_1^2 + x_2^2$
We can say this because $x_1^2$ will be non-negative since it's a square. Then, take the square root of both sides:
$\sqrt{x_2^2} \leq \sqrt{x_1^2 + x_2^2} \implies |x_2| \leq \sqrt{x_1^2 + x_2^2}$
We can then multiply both sides of the right inequality above by $|u|$ to obtain:
$|x_2||u| \leq |u|\sqrt{x_1^2 + x_2^2} = |u|||\vec{x}||$
Case 1: $x_2, u \geq 0$
In this case, we are done since $|x_2||u| = x_2u$
Case 2: $x_2, u < 0$
We simply notice that $|x_2||u| = (-x_2)(-u) = (-1)^2(x_2)(u) = x_2u$ and we are done.
Case 3: $x_2 \geq 0, u < 0$ (Also covers the case where $u \geq 0, x_2 < 0$)
Notice that $|x_2||u| = x_2(-u) \geq 0$
Multiplying both sides of this inequality by $-1$ obtains:
$x_2u \leq 0 \implies x_2u \leq x_2(-u) \leq |u|||\vec{x}||$