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.

Hint:
The discriminate of $$x^2+4x+5=0$$ is negative so the sign of $x^2+4x+5$ is the sign of the coefficient of $x^2$ which is $+1$.
