Prove existence of $c$ such that $f(x)\ge c g(x)$ Let $f$ and $g$ be non-negative continuous functions on $[0,1]$ such that $f(x)>g(x)$ for all $x$ in $[0,1]$. Show that there exists a constant $c>1$ such that for all $x$ in $[0,1]$ we have $f(x)\ge c g(x)$.
I tried using the fact that since $[0,1]$ is compact and $f$ and $g$ are continuous $f$ and $g$ each attain a maximum and a minimum on $[0,1]$, but it didn't get me anywhere. Anyone know how to prove this?
 A: Take $h(x)=\frac{g(x)}{f(x)}$. This function is continuous and well-defined because $f(x)>g(x)\geq 0$.
But the above says that $h(x)<1$ for all $x$. Then $h$ has a maximum, $m$ and it is less than $1$. We have that $\frac{g(x)}{f(x)}\leq m$ and therefore $m\cdot f(x)\geq g(x)$. Take $c=\frac1m$ then $c>1$ and we have $f(x)\geq c\cdot g(x)$.
A: Since $f(x) >g(x) \geq 0$ you know that $f(x) \neq 0$.
Then the function $\frac{g(x)}{f(x)}$ is continuous on $[0,1]$ and thus It attains it's maximum.
Let $N$ be its maximum, and let $x_0$ be the point where it is attained. Then
$$N=\frac{g(x_0)}{f(x_o)} <1$$
and since $N$ is the maximum
$$f(x)\geq \frac{1}{N} g(x) \,.$$
A: Edit: The function $f(x)-g(x)$ is continuous on $[0,1]$ and strictly positive, so it attains a minimum $k_1>0$.  The function $g(x)$ is also continuous on this interval, and attains a maximum $k_2\geq 0$. If $k_2=0$, then $g(x)=0$ and the problem is trivial, so assume that $k_2>0$.  With this maximum and minimum in mind, consider the function 
$$h(x)=f(x)-\left(1+\frac{k_1}{2k_2}\right)g(x).$$
What can we say about it's minimum on $[0,1]$?
A: Set  $\epsilon>0$ and 
$$
C=\left(1+\epsilon\right)>0
$$
Note that $\epsilon\cdot g(x)>0$ and
\begin{align}
f(x)-C\cdot g(x)=
&
f(x)-\left(1+\epsilon\right)\cdot g(x)
\\
=
&
f(x)-g(x)+\epsilon\cdot g(x)
\\
>
&
\epsilon\cdot g(x)
\\
>
&
0
\end{align}
A: Hint: 
Let $A = \{x \mid g(x) \neq 0\}$. Set $c = \inf_{x \in A} \frac{f(x)}{g(x)}$.
