Proving one function is greater than another How can I prove $f(x)$ $>$ $g(x)$ for all $x > 0$ given $f(x) = (x+1)^{2}$ and $g(x) = 4qx$ where $q$ is a constant in $(0, 1)$?
My approach was to show that $(x+1)^2 > 4qx$ for the interval endpoints, e.g. $q=0$ and $q=1$. E.g. $(x+1)^2 \geq 4x$ for all $x$ and $(x+1)^2 > 0$ for all $x$. However, $q \neq 0,1$ so $f(x) > g(x)$ for all $x$. However, I'm looking for something more mathematically rigorous. Any suggestions?

$\begin{align}
f(x)-g(x)
&=(x+1)^2-4qx\\
&=x^2+2x+1-4qx\\
&=x^2+2(1-2q)x+1\\
&=x^2+2(1-2q)+(1-2q)^2-(1-2q)^2+1\\
&=(x+1-2q)^2+1-(1-2q)^2\\
\end{align}
$.
Since $0 <q < 1$,
$-1 < 1-2q < 1$
so $0 < (1-2q)^2 < 1$
so $1 > 1-(1-2q)^2 > 0$
so (finally)
$f(x) > g(x)$.
Another way to get this final inequality is
$1-(1-2q)^2 
= (1-(1-2q))(1+(1-2q))
= 2q(2-2q)
= 4q(1-q)
$
and both $1$ and $1-q$ are positive.
