Show that $\|f_1+f_2\| \leq \|f_1\| + \|f_2\|$ using Minkowski's inequality I am trying to show that: $\|f_1+f_2\| \leq \|f_1\| + \|f_2\|$ using the Minkowski inequality.
for: 
$$ \|f\| = \left(\int_0^1 \left[|f|^2 + |f'|^2\right]\ dx\right)^{1/2}.$$
I dont see how I can apply Minkowski since $\|f\|$ is not in a form that I would think Minkowski could be applied? 
As always, thanks in advance!
 A: All you need to do is a little algebra to substitute $f_1 + f_2$ into the new definition. 
To apply Minkowski inequality, let's denote the traditional $L_2$ norm of $f$: $\left(\int_0^1 |f|^2 dx\right)^{1/2}$ by $\|f\|_0$. We have:
\begin{align}
& \|f_1 + f_2\|^2 \\
= & \int_0^1 |f_1 + f_2|^2 dx + \int_0^1 |f_1' + f_2'|^2 dx \\
= & \|f_1 + f_2\|_0^2 + \|f_1' + f_2'\|_0^2 \\
\leq & (\|f_1\|_0 + \|f_2\|_0)^2 + (\|f_1'\|_0 + \|f_2'\|_0)^2 \quad \text{we apply Minkowski inequality here}.\\
= & \|f_1\|_0^2 + \|f_2\|_0^2 + \|f_1'\|_0^2 + \|f_2'\|_0^2 + 2\|f_1\|_0\|f_2\|_0 + 2\|f_1'\|_0\|f_2'\|_0 \\
= & \int_0^1 |f_1|^2 dx + \int_0^1 |f_2|^2 dx + \int_0^1 |f_1'|^2 dx + \int_0^1 |f_2'|^2 dx \\
& + 2\sqrt{\int_0^1 |f_1|^2 dx\int_0^1 |f_2|^2 dx} + 2\sqrt{\int_0^1 |f_1'|^2 dx\int_0^1 |f_2'|^2 dx} \\
= & \|f_1\|^2 + \|f_2\|^2 + 2\sqrt{\int_0^1 |f_1|^2 dx\int_0^1 |f_2|^2 dx} + 2\sqrt{\int_0^1 |f_1'|^2 dx\int_0^1 |f_2'|^2 dx}
\end{align}
So to show the result, it suffices to show that
$$2\sqrt{\int_0^1 |f_1|^2 dx\int_0^1 |f_2|^2 dx} + 2\sqrt{\int_0^1 |f_1'|^2 dx\int_0^1 |f_2'|^2 dx} \leq 2\|f_1\|\|f_2\| \tag{1}.$$
For simplicity, denote $\int_0^1 |f_i|^2 dx$ by $A_i$, $\int_0^1 |f_i'|^2 dx$ by $B_i$, $i = 1, 2$, then $(1)$ can be rewritten as:
$$\sqrt{A_1A_2} + \sqrt{B_1B_2} \leq \sqrt{A_1 + B_1}\sqrt{A_2 + B_2},$$
which is straightforward to verify by AM-GM inequality. The proof is complete.
