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?
 A: Related to one of the above solutions, we know that there are positive values to $h(x)=(x+1)^2-4qx$ (why?) and so if there were also non-positive values, there would be roots to $h(x)=0$. But $h(x)=x^2+(2-4q)x+1$ so $h(x)=0$ has real roots if and only if $(2-4q)^2-4\geq 0$, which means $(1-2q)^2\geq 1$.  But if $q\in(0,1)$, $1-2q\in(-1,1)$ and thus $(1-2q)^2<1$.
A: Since $q \in (0,1)$ and $x > 0$, we know $4xq < 4x$, so it suffices to prove the stronger claim $(x+1)^2 \geq 4x$ for all $x > 0$. The latter is equivalent to showing $(x+1)^2 - 4x \geq 0$. To that end, we have
\begin{align*}
(x+1)^2 - 4x &= x^2 + 2x + 1 - 4x\\
&= x^2 - 2x + 1\\
&= (x - 1)^2\\
&\geq 0.
\end{align*}
A: Try this: $${ x }^{ 2 }+2x+1-4xq\ge 0\\ { x }^{ 2 }-2x\left( 2q-1 \right) +1+{ \left( 2q-1 \right)  }^{ 2 }-{ \left( 2q-1 \right)  }^{ 2 }\ge 0\\ { \left( x-2q+1 \right)  }^{ 2 }\ge { \left( 2q-1 \right)  }^{ 2 }-1$$
You can use this for any $q$
A: $\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.
