<s>[INST] Proof of $(\forall x)(x^2+4x+5 \geqslant 0)$ $(\forall x)(x^2+4x+5\geqslant 0)$ universe is $\Re$
I went about it this way
$x^2+4x \geqslant -5$
$x(x+4) \geqslant -5$
And then I deduce that if $x$ is positive, then $x(x+4)$ is positive, so it's $\geqslant 5$
If $ 0 \geqslant x \geqslant -4$, then $x(x+4)$ is also $\geqslant -5$.
If $ x < -4$, then $x(x+4)$ will be negative times negative = positive, so obviously $\geqslant -5$
I'm wondering if there's an easier to solve this problem? My way seems a little clunky.
 [/INST] \begin{align}
x^2+4x+5 
&= x^2 + 2x(2) +(2)^2 +1\\
&= (x+2)^2 +1
\end{align}
$(x+2)^2\ge0$ for any $x\in\mathbb{R}$. Adding $1$ for both sides of inequalities, we have
$$
(x+2)^2+1\ge 1$$
for any $x\in \mathbb{R}$.
As $x^2+4x+5=(x+2)^2+1$,
$$
x^2+4x+5\ge 1>0$$
for any $x\in \mathbb{R}$.
Finally,
$$
x^2+4x+5>0$$
for any $x\in \mathbb{R}$.
Note
The quadratic expression $x^2+4x+5$ is always greater than zero, it never equals to zero. Be careful!
</s>