Integral Inequality Proof Using Hölder's inequality I'm working on the extra credit for my Calculus 1 class and the last problem is a proof. We have done proofs before, but I'm unsure of how to approach this problem. Any help would be much appreciated, I'm just not entirely sure how to start. Specifically, showing how the continuous function defined by the problem is to taken into account. I understand that it defines the bounds for the integrals, but beyond that I am unsure of how they are affected. My intuition tells me that it implies the use of a specific theorem, but no specific theorem is apparent to me from simple observation. I've found evidence that Holder's Inequality would be helpful, but I'm not certain as to how to apply this. The problem is as follows:

Show that for any continuous function $f:[-1,1]\to\mathbb R$ we have that:
$$2\int_{-1}^{1} f^2(x) dx \ge \left(\int_{-1}^{1}f(x)dx\right)^2+3\left(\int_{-1}^{1}xf(x)dx\right)^2$$

I've tried multiple approaches essentially yielding nothing though. My understanding of how to manipulate integrals in a general manner is weak at best. I did get a hint from my TA that said the proof is almost a page long and uses results beyond Calculus 1, so technically I shouldn't even be able to do this proof, but it is worth a significant amount of my grade and I could really use all the help I can get. We actually only started integrals about 3 weeks ago. Not to mention it has now started to bother me that I can't solve it despite spending so much time. The extra credit is due Friday, but my Final is tomorrow, so I won't be able to spend much more time on it.
Thanks for any help you guys can provide. 
 A: For brevity, write $\int F $ for $ \int_{-1}^{1} F(x) \, dx $.
Let $\int f = A$. Then
$$ \int \left( f-\tfrac{1}{2}A \right) = \int f - A = 0 $$
Now use Cauchy–Schwarz:
$$ 3\left( \int x \left(f-\tfrac{1}{2}A \right) \right)^2 \leqslant 3\left( \int x^2 \right) \left( \int \left(f-\tfrac{1}{2}A \right)^2 \right) = 2 \left( \int f^2 - A\int f + \frac{A^2}{4} \int 1 \right) \\
= 2\left( \int f^2 - A^2 +\frac{A^2}{2} \right) = 2\int f^2 -A^2 $$
Also, we have
$$ 3\left( \int x(f-A) \right)^2 = 3\left(\int xf\right)^2, $$
because $\int x = 0 $, and so we are done, after rearranging.
(Actually, you don't need $f$ to be continuous, just that all three of the integrals exist.)

Now, I shall do you one better: there is a set of polynomials $P_n(x)$, where $P_n$ is of degree $n$, such that you can find a similar inequality in the same way: for every positive integer $N$, if $\int x^n f$ exists for $0 \leqslant n \leqslant N$, there is an inequality
$$ 2\int f^2 \geqslant \sum_{n=0}^N (2n+1)\left( \int P_n f \right)^2 $$
If we take $N=1$, then this is for $P_0(x)=1$, $P_1(x)=x$. The key to the proof is the orthogonality of the $P_n$: we have
$$ \int P_n P_m = \begin{cases} \frac{2}{2n+1} & n = m \\ 0 & n \neq m \end{cases}. $$
We can find unique $P_n$ with this property, given only that $P_n$ has degree $n$. These are called the Legendre polynomials, probably the most basic example of orthogonal polynomials.
