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.
 A: Just note that $(x+2)^2=x^2+4x+4$.
A: Assuming  $x$ is real, $(x+2)^2\ge0\implies x^2+4x+4\ge0$
A: 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 $+1$.
A: Take a look at the discriminant of your quadratic which is 
$$ 16-4(1)*5 <0 $$ Since it is negative your equation will be greater than zero with strict inequality for all $x$
A: \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!
A: One of the best ways in general to handle questions about quadratics is via the process of completing the square: that is, turning a quadratic of the form $P(x) = x^2+bx+c$ into one of the form $Q(x) = (x+r)^2+s$.  To do this, note that the latter can be written as $x^2+2rx+r^2+s$, and we can equate terms; this means that $r$ must be equal to $\frac b2$, and then since $s+r^2=c$, we must have $s=c-r^2=c-\frac{b^2}{4}$ (and if you were to use an 'a' term for your quadratic and pursued the algebra along this path a little further, you'd derive the famous quadratic formula).
In the case at hand, we have $b=4$ so we know that $r=\frac b2=2$, and we get that $x^2+4x+5$ = $(x+2)^2+s$ = $x^2+4x+4+s$, so $s=1$ and the quadratic can be written in the form $(x+2)^2+1$.  In this form it's immediate that it's positive for all $x$ (since $(x+2)^2\geq 0$ for all $x$ and $1\gt 0$).  What's more, we can see that its minimum value is acheived where $(x+2)^2$ is at a minimum - that is, when $x+2=0$, or in other words when $x=-2$; thus, the vertex of the parabola given by $y=P(x) = x^2+4x+5$ is at $(x,y) = (-2, P(-2)) = (-2, 1)$.
