Proof

By hypothesis, $p,q\in (1,\infty)$ with $\frac 1p+\frac 1q=1$ and $x=(x_1,x_2,\ldots x_n)$ and $y=(y_1,y_2,\ldots y_n)$ are vectors of a vector space $V$ over the field of real numbers $\mathbb R$ or the field of complex numbers $\mathbb C$.
Without loss of generality we can assume that the p-norms $||x||_p|$ and $||y||_q$ are both $\neq 0$, otherwise the inequality to be proven is trivial.
For $\nu=1,\ldots,n$ we set $$\xi_\nu:=\left(\frac{|x_\nu|}{||x||_p}\right)^p,\quad\eta_\nu:=\left(\frac{|y_\nu|}{||y||_q}\right)^q.$$
Note that by this definition and the definition of p-norms, $\sum_{\nu=1}^n\xi_\nu=1$ and $\sum_{\nu=1}^n\eta_\nu=1$ and $\xi_\nu,\eta_\nu\in(0,1)$ for $\nu=1,\ldots,n.$
Therefore, we can apply the lemma about an upper bound for the product of general powers and get for $\nu=1,\ldots,n$ the inequalities $$\frac{|x_{\nu} y_{\nu}|}{||x||_p ||y||_q}=\xi_{\nu}^{1/p}\eta_{\nu}^{1/q}\le \frac{\xi_{\nu}}{p}+\frac{\eta_{\nu}}{q}.$$
Summing up both sides for $\nu=1,\ldots,n$ yields $$\frac{1}{||x||_p ||y||_q}\sum_{\nu=1}^n |x_{\nu} y_{\nu}|\le \frac 1p+\frac 1q=1.$$
It follows $$\sum_{\nu=1}^n|x_\nu y_\nu|\le ||x||_p||x||_q.$$
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