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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?

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Yes, seems right. You would like to add. $(x+1)^2 \ge 4x$ because $(x-1)^2 \ge 0$ –  Inceptio May 20 '13 at 5:01
    
you can derive both functions in order to get $f\prime(x)=2x+2,g'(x)=4q$ and since $0\leq q \leq 4$ exist $x_0$ such that $\forall x>X_0, 2x-2>4q$. You also know that in x=0, f(x)=1 while g(x)=0. That proves your claim. –  Coargu Aliquis May 20 '13 at 5:07
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4 Answers

up vote 5 down vote accepted

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*}

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Very short and elegant. (+1) –  Bryan Urízar May 20 '13 at 7:17
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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 $(1-2q)^2<1$.

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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 $q$

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Yes I see now, thanks. –  nikamed May 20 '13 at 5:15
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$\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.

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