I'm having trouble establishing these inequalities without using Cauchy-Bunyakovsky-Schwarz or Minkowski inequalities. Does anyone have a helpful hint?
$f$ and $g$ are continuous on $[0,1]$:
$$ (a) \hspace{5mm} \Bigg(\int_{0}^{1} f(x)g(x)dx\Bigg)^{2} \leq \int_{0}^{1} f(x)^{2}dt \int_{0}^{1}g(x)^{2}dt$$
$$ (b) \hspace{5mm} \Bigg(\int_{0}^{1}(f(x) + g(x))^{2}dt\Bigg)^{\frac{1}{2}} \leq \Bigg(\int_{0}^{1}f(x)^{2} dt\Bigg)^{\frac{1}{2}} + \Bigg(\int_{0}^{1} g(x)^{2}dt\Bigg)^{\frac{1}{2}}$$
Hint for (a): Consider that
$$\int_0^1 \int_0^1 [f(x)g(y) - g(x)f(y)]^2\, dx\, dy \ge 0.$$
Expand the double integral above.
Hint for (b): Write
$$\int_0^1 [f(x) + g(x)]^2\, dx = \int_0^1 [f(x) + g(x)]f(x)\, dx + \int_0^1 [f(x) + g(x)]g(x)\, dx.$$
Then by the triangle inequality,
$$\left|\int_0^1 [f(x) + g(x)]^2\, dx\right| \le \left|\int_0^1 [f(x) + g(x)]f(x)\, dx\right| + \left|\int_0^1 [f(x) + g(x)]g(x)\, dx\right|.$$
Use part (a) to show that the first integral on the right-hand side is less than or equal to
$$ \left(\int_0^1 [f(x) + g(x)]^2\, dx\right)^{1/2}\left(\int_0^1 [f(x)]^2\, dx\right)^{1/2}$$
and that the second integral is less than or equal to
$$\left(\int_0^1 [f(x) + g(x)]^2\, dx\right)^{1/2}\left(\int_0^1 [g(x)]^2\, dx\right)^{1/2}.$$