I am trying to show that $||x+y||_p \leq ||x||_p + ||y||_p$ where $p$ is an integer larger than 1, but not infinity (I proved those cases already), and $||x||_p = (\sum_{i=1}^n |x_i|^p)^{\frac{1}{p}}$
We were also given a hint:
Holder's inequality states $\langle u,w \rangle \leq ||u||_p ||w||_q$ when $\frac{1}{p}+\frac{1}{q} = 1$ implies $q = \frac{p}{p-1}$ and $\langle u,w \rangle = \sum_{i=1}^n u_iw_i$
I'd appreciate any help offered here...I already showed its true for $p=1$ and $p= \infty$ as single cases.
Since the cases $p = 1$ and $p = \infty$ are already done, I treat only $1 < p < \infty$.
For arbitrary $a,b\in\mathbb{C}$, we have
$$\lvert a+b\rvert^p = \lvert a+b\rvert\cdot\lvert a+b\rvert^{p-1} \leqslant \lvert a\rvert\cdot\lvert a+b\rvert^{p-1} + \lvert b\rvert\cdot \lvert a+b\rvert^{p-1}.$$
Thus
$$\begin{align} \lVert x+y\rVert_p^p &= \sum_{i=1}^n \lvert x_i + y_i\rvert^p\\ &\leqslant \sum_{i=1}^n \underbrace{\lvert x_i\rvert}_{u_i}\cdot \underbrace{\lvert x_i+y_i\rvert^{p-1}}_{w_i} + \sum_{i=1}^n \underbrace{\lvert y_i\rvert}_{u_i} \cdot \underbrace{\lvert x_i+y_i\rvert^{p-1}}_{w_i}. \end{align}$$
Applying Hölder's inequality with $q = \frac{p}{p-1}$ to each of the last sums, we obtain
$$\begin{align} \lVert x+y\rVert_p^p &\leqslant \left(\sum_{i=1}^n \lvert x_i\rvert^p\right)^{1/p} \left(\sum_{i=1}^n \left(\lvert x_i+y_i\rvert^{p-1}\right)^{p/(p-1)}\right)^{(p-1)/p}\\ &\qquad + \left(\sum_{i=1}^n \lvert y_i\rvert^p\right)^{1/p} \left(\sum_{i=1}^n \left(\lvert x_i+y_i\rvert^{p-1}\right)^{p/(p-1)}\right)^{(p-1)/p}\\ &= \lVert x\rVert_p \left(\sum_{i=1}^n \lvert x_i+y_i\rvert^p\right)^{(p-1)/p} + \lVert y\rVert_p \left(\sum_{i=1}^n \lvert x_i+y_i\rvert^p\right)^{(p-1)/p}\\ &= \lVert x\rVert_p \lVert x+y\rVert_p^{p-1} + \lVert y\rVert_p\lVert x+y\rVert_p^{p-1}\\ &= \left(\lVert x\rVert_p + \lVert y\rVert_p\right) \cdot\lVert x+y\rVert_p^{p-1}. \end{align}$$
Now, either $\lVert x+y\rVert_p = 0$, in which case the inequality $\lVert x+y\rVert_p \leqslant \lVert x\rVert_p + \lVert y\rVert_p$ trivially holds, or we can divide the last inequality we obtained by $\lVert x+y\rVert_p^{p-1}$ to obtain the desired triangle inequality.