Prove if $f:[a,b]\rightarrow \mathbb{R}$ is continous and not everywhere zero, then $\int_{a}^{b}f^2(x)dx>0.$ Claim:

If $f:[a,b]\rightarrow \mathbb{R}$ is continuous and not everywhere zero, then $\int_{a}^{b}f^2(x)dx>0.$

I have the following theorem that seems applicable here - more specifically the contrapositive:

If $g$ is a continuous nonnegative function on $[a,b]$ and if $\int_{a}^bg=0$, then $g=0$ on $[a,b]$.

The contraposition to this statement being (?):

If $g\not=0$ on $[a,b]$ and $g$ is a nonnegative function on $[a,b]$, then $\int_{a}^bg\not=0$. Where $g\not=0 \Leftrightarrow g>0$, since $g$ is assumed to be nonnegative.


I'm not sure if the theorem posted above is at all applicable. However, below are my ideas to proceed.
First note that there exists at least one (?) $x_0\in[a,b]$ s.t. $f(x_0)>0$ or $f(x_0)<0$. I'm inclined to think that there is some neighborhood about $x_0$ so that we can preserve continuity of $f$ on $[a,b]$. 
Once that neighborhood is established, we may consider cases: 


*

*If $f(x)>0$, then $f^2(x)>0$, $\forall x\in V_{\varepsilon}(x_0)$

*If $f(x)<0$, then $f^2(x)>0$, $\forall x\in V_{\varepsilon}(x_0)$


If the above is correct, I'm not sure how to make the jump that in either case $\int_a^bf^2(x)dx>0$ is necessarily true. 
This is a homework problem, so I'd prefer just a hint. 
 A: An integrable function doesn't have to be continuous, the theorem you wrote is false (the assumption of continuity is missing or there should be "almost everywhere on $[a,b]$").

If a nonnegative continuous function $g$ is not 0 everywhere, then it is positive in some point and - by continuity - positive on some interval $[c,d]\subset[a,b]$. Then $\int_a^b g(x)dx\ge\int_c^df(x)dx>0$.
For any continuous function $f$, the function $g=f^2$ is nonnegative and continuous, so the above theorem works fine.
A: You're on the right track.  Here are a few suggestions to keep it going:
1) You have an interval where you can bound your function away from zero.  Can you use the same technique to bound it above some positive number, say $\frac{f(x_0)}{2}$?
2) Once you have such an interval, can you break the integral over $[a, b]$ into three intervals, the middle one of which you have just learned something considerable about?
A: Hint: You are on the right track. To move on, denote that neighborhood as $(x_0 - \varepsilon, x_0 + \varepsilon)$, such that for all $x$ in this neighborhood, $|f(x) - f(x_0)| < |f(x_0)|/2$ so that $|f(x)| > |f(x_0)|/2$. Then bound the integration by 
$$\int_{x_0 - \varepsilon}^{x_0 + \varepsilon} f^2(x) dx.$$
See what you will get then.
A: Note that $f^{2}(x) \geq 0$ everywhere. Therefore, if $a \leq c < d \leq b$, i.e. $(c, d)$ is a subinterval of $(a, b)$, then $\int_{a}^{b} f^{2}(x) \mathrm{d}x = \int_{a}^{c} f^{2}(x) \mathrm{d}x + \int_{c}^{d} f^{2}(x) \mathrm{d}x + \int_{d}^{b} f^{2} (x) \mathrm{d}x \geq \int_{c}^{d} f^{2}(x) \mathrm{d} x$. So let $p$ be a point on $(a, b)$ such that $f^{2}(p) = q > 0$, i.e. $p$ is some point where $f^{2}(x)$ is not $0$. Then by the definition of continuity, for $\epsilon = q / 2$ exists some $\delta > 0$ such that if $|x - p| < \delta$, then $q - \epsilon < f(x) < q + \epsilon$, which is equivalent to saying that if $p - \delta < x < p + \delta$, then $q / 2 < f(x) < 3q / 2$. Let $c = p - \delta, d = p + \delta$. Then since $f^{2}(x) > q / 2$ on $(c, d)$, it follows that $\int_{c}^{d} f^{2}(x) \mathrm{d}x \geq (d - c) (q / 2) = \delta q > 0$. Thus by our earlier observation that $\int_{a}^{b} f^{2}(x) \mathrm{d}x \geq \int_{c}^{d} f^{2}(x) \mathrm{d}x$, we have that $\int_{a}^{b} f^{2}(x) \mathrm{d}x > 0$.
