$C_{1}f(x) \leq g(x) \leq C_{2}f(x)$ where $f,g $ be continuous and positive functions 
Let $f$ and $g$ be continuous and positive functions on $[a,b]$. Then there are positive $C_{1},C_{2}$ such that
$$
C_{1}f(x) \leq  g(x) \leq C_{2}f(x).
$$

I know that there are $\alpha_{1},\alpha_{2} \in [a,b]$ such that
$$
f(\alpha_1) \leq f(x) \leq f(\alpha_2), \ \ \forall x \in [a,b].
$$
And
there are $\beta_{1},\beta_{2} \in [a,b]$ such that
$$
g(\alpha_1) \leq g(x) \leq g(\alpha_2), \ \ \forall x \in [a,b].
$$
I'm also trying to follow by contradiction: assuming there is $x_{0} \in [a,b]$
$$
g(x_0) > Af(x_0), \forall A > 0.
$$
But I can't get anywhere
 A: Consider $h(x) = \frac{g(x)}{f(x)}$. Then as $f,g$ are both positive on $[a,b]$ and $[a,b]$ is closed, $f(x)$ is bounded and bounded away from $0$, and likewise with $g(x)$; thus the function $h(x)$ is well-defined, positive, and bounded on $[a,b]$.

*

*Thus, there are positive real numbers $C_1$ and $C_2$ such that $C_1 \le h(x) \le C_2$ for all $x \in [a,b]$.


*Furthermore, for all $x$ on $[a,b]$ the equation $g(x) =h(x)f(x)$ holds.
Thus putting 1. and 2. together yields:
$$\forall x \in [a,b]: \quad C_1f(x) \le h(x)f(x) = g(x) \le C_2f(x),$$
where $C_1$ and $C_2$ are as defined in 1., and the equation $h(x)f(x)=g(x)$ for all $x \in [a,b]$ was as observed in 2. This is what we want.
***If we weaken these conditions even somewhat however, then there are counterexamples. Suppose instead of an interval of the form $[a,b]$ we consider an interval of the form $(a,b]$. Then consider the interval $(0,1]$, with $f(x) = \frac{1}{x}$, and $g(x)=1$ for all $x \in (0,1]$. Then here there is no such positive $C_1$.
A: By Heine Borel theorem, $[a,b]$ is a compact set. Since $f,g$ are continuous function, by extremal value theorem, the $\underset{x\in[a,b]}{\max}$,$\underset{x\in[a,b]}{\min}$ of f and g exists. So there always exists two constant $C_1$ and $C_2$ s.t $$C_1\underset{x\in[a,b]}{\max}f(x)\le\underset{x\in[a,b]}{\min} g(x)$$ and $$\underset{x\in[a,b]}{\max}g(x)\le C_2 \underset{x\in[a,b]}{\min} f(x)$$
Thus we have $$C_1 f(x)\le g(x) \le C_2 (x)$$
