I have a problem with one exercise. I have to prove that $L^2$ space is Hilbertian. So I think that the best way is to check out inner product by definition of norm, so:
\begin{equation*} (f,g)=\int\limits_{-\infty}^{\infty}fg^{*}d^3r \end{equation*}
That was obvious to show that (if $||f||=\sqrt[2]{(f,f)}$):
1) $||f||=0 \iff f=0$
2) $||a \cdot f||=|a| \cdot ||f||$
But I have aproblem with the triangle inequality: $\|f+g\|\leq\|f\|+\|g\|$.
Any sugestions?
The triangle equality is true in general for any inner product space, and the defining properties of inner product spaces are pretty easy to see in this case (it looks like you already have most of them).
The proof of the triangle inequality goes through Cauchy-Schwarz, i.e. $|(f,g)| \leq ||f||\cdot||g||$. To see Cauchy-Schwarz, observe that by renormalizing (and since it is trivial if either function has norm 0), we can without loss of generality assume $||f|| = ||g|| = 1$. We are then trying to show $|(f,g)| \leq 1$. Choose $\mu$ with $|\mu| = 1$ so that $(\mu f,g)$ is real and non-negative. Observe: \begin{align*} 0 &\leq (\mu f - g, \mu f - g) \\ &= ||\mu f||^2 + ||g||^2 - (\mu f, g) - (g, \mu f) \\ &= 2 - (\mu f,g) - \overline{(\mu f,g)} \\ &= 2 - 2(\mu f, g)\text{.} \end{align*} Therefore \begin{align*} 1 &\geq (\mu f,g) \\ &= |(\mu f,g)| \\ &= |\mu|\cdot|(f,g)| \\ &= |(f,g)|\text{.} \end{align*} We've thus shown Cauchy-Schwarz.
To see the triangle inequality from Cauchy-Schwarz, observe: \begin{align*} ||f + g||^2 &= (f+g,f+g) \\ &= ||f||^2 + ||g||^2 + (f,g) + \overline{(f,g)} \\ &\leq ||f||^2 + ||g||^2 + 2|(f,g)| \\ &\leq ||f||^2 + ||g||^2 + 2||f||\cdot||g|| = (||f|| + ||g||)^2 \text{,} \end{align*} as desired.