How to show that $\left({\int_a^bfg}\right)^2 ≤ \int_a^b f^2 \int_a^bg^2$? 
Let $f$ and $g$ be continuous on $[a,b]$ Show that
  $$\left({\int_a^bfg}\right)^2 ≤ \int_a^b f^2 \int_a^bg^2$$


Okay, the first integral on the left is squared by the way.
I'm trying to show that this equality holds, but couldn't really do anything.
I'm given this hint :
$$\int_a^b(f+\lambda g)^2 ≥0 $$
 A: This is the Cauchy—Schwarz inequality (or at least one instantiation of it). Here is a standard and elementary derivation, based on the hint you are given.
Start with your hint: fix any two continuous (and thus integrable) real functions $f,g$ on $[a,b]$. For every $\lambda\in\mathbb{R}$,
$$
0 \leq \int_a^b (f+\lambda g)^2 = \int_a^b f^2 + 2\lambda\int_{a}^b fg + \lambda^2 \int_a^b g^2
$$
which is a quadratic polynomial in $\lambda$: for all $\lambda$,
$$0 \leq C + \lambda B + \lambda^2 A$$
where $A,B,C$ are given by the integrals.
Since the quadratic polynomial is always non-negative, it must be the case that its determinant is non-positive:
$$
\Delta = B^2- 4AC \leq 0.
$$
Replacing $A,B,C$ by their value gives:
$$
4\left(\int_a^b fg\right)^2 - 4\int_a^b f^2\int_a^b g^2 \leq 0.
$$
which, after rearrangement, is the inequality you want to prove.
A: According to Cauchy-Schwarz inequality, $$\int_a^bfg\,\mathrm{d}x\le\left(\int_a^bf^2\,\mathrm{d}x\right)^{1/2}\left(\int_a^bg^2\,\mathrm{d}x\right)^{1/2}$$ From here we can derive the desired inequality. 
For detail see Cauchy-Schwarz Inequality.
