Let \((x_n)_{n \in \mathbb{N}}, (y_n)_{n \in \mathbb{N}}, (z_n)_{n \in \mathbb{N}}\subseteq \mathbb R\) be sequences such that \(x_n \leq y_n \leq z_n\) for all \(n \in \mathbb{N}\). If \(\lim_{n \to \infty}x_n=\lim_{n \to \infty}z_n=L\in\mathbb R\), then \(\lim_{n \to \infty}y_n=L\).
Every bounded and monotonic sequence in \(\mathbb R\) converges to some \(x\in \mathbb R\).
Binomial theorem. For every \(n \in \mathbb{N}\) and \(x,y \in \mathbb{R}\) one has \[(x+y)^n=\sum_{k=0}^n {n \choose k}x^ky^{n-k}, \quad \text{ where } \quad {\color{red}{n \choose k}=\frac{n!}{k!(n-k)!},}\] and \(n!=1 \cdot 2 \cdot 3 \cdot \ldots \cdot n\), for all \(n \in \mathbb{N}\) and \(0!=1\).
Theorem.
If \(p>0\), then \(\lim_{n \to \infty}\frac{1}{n^p}=0\).
If \(p>0\), then \(\lim_{n \to \infty}\sqrt[n]{p}=1\)
\(\lim_{n \to \infty}\sqrt[n]{n}=1\).
If \(p>0\) and \(\alpha \in \mathbb{R}\), then \(\lim_{n \to \infty}\frac{n^{\alpha}}{(1+p)^n}=0\).
If \(|x|<1\), then \(\lim_{n \to \infty}x^n=0\).
Proof of (a): Take \(\varepsilon>0\) be arbitrary, but fixed. Then \[n>\left(\frac{1}{\varepsilon}\right)^{1 / p},\] which is possible by the Archimedian property.
Proof of (b): If \(p>1\) set \(x_n=\sqrt[n]{p}-1\), then \(x_n>0\) and by Bernoulli’s inequality \[1+nx_n \leq (1+x_n)^{n}=p,\]
so that
\[0<x_n \leq \frac{p-1}{n}.\]
But \[\lim_{n \to \infty}\frac{p-1}{n}=0,\]
thus by the squeeze theorem we conclude \[\lim_{n \to \infty}x_n=0\] as desired.
Proof of (c): Set \(x_n=\sqrt[n]{n}-1\). Then \(x_n \geq 0\) and by the binomial theorem \[n=(1+x_n)^n \geq {n \choose 2}x_n^2=\frac{n(n-1)}{2}x_n^2.\]
Hence \[0 \leq x_n \leq \left(\frac{2}{n-1}\right)^{1 / 2} \quad \text{ for } \quad n \geq 2.\]
But \[\lim_{n \to \infty}\left(\frac{2}{n-1}\right)^{1 / 2}=0.\] Thus by the squeeze theorem \[\lim_{n \to \infty}x_n=0\] as desired. $$\tag*{$\blacksquare$}$$
Proof of (d): Let \(k \in \mathbb{N}\) so that \(k>\alpha\). For \(n>2k\) by the binomial theorem \[(1+p)^n>{n \choose k}p^k=\frac{n(n-1)\ldots (n-k+1)}{k!}p^k> \frac{n^k p^k}{2^k k!},\]
since \(n \geq \frac{n}{2}, n-1 \geq \frac{n}{2},\ldots, n-k+1 \geq \frac{n}{2}\). Hence \[0<\frac{n^{\alpha}}{(1+p)^n}<\frac{2^kk!}{p^k}n^{\alpha-k} \quad \text{ for } \quad n > 2k.\]
Since \(\alpha-k<0\) thus \(\lim_{n \to \infty}n^{\alpha-k}=0\) by (a) and by the squeeze theorem \(\lim_{n \to \infty}\frac{n^{\alpha}}{(1+p)^{n}}=0\). $$\tag*{$\blacksquare$}$$
Proof of (e): Take \(\alpha=0\) in (d) and observe that if \(0<x<1\) then the sequence \(x_n=x^n\) is decreasing and bounded. Thus \(\lim_{n \to \infty}x_n=0\). $$\tag*{$\blacksquare$}$$
Proposition. If \(a>0\) and \(\lim_{n \to \infty}x_n=x_0\), then \(\lim_{n \to \infty}a^{x_n}=a^{x_0}\).
Proof. It suffices to prove that \(\lim_{n \to \infty}a^{x_n}=1\) if \(\lim_{n \to \infty}x_n=0\).
Assume \(a>1\). By the previous theorem we know that \[\lim_{n \to \infty}a^{1 / n}=\lim_{n \to \infty}a^{-1 / n}=1.\] Thus for any \(\varepsilon>0\) there is \(M_{\varepsilon} \in \mathbb{N}\) such that for any \(m \geq M_{\varepsilon}\) \[1-\varepsilon < a^{-1 / m}<a^{1 / m}<1+\varepsilon.\]
Now since \(\lim_{n \to \infty}x_n=0\) we find \(N_{m,\varepsilon}\in \mathbb{N}\) so that for \(n \geq N_{\varepsilon,m}\) \[|x_n|<\frac{1}{m} \iff -\frac{1}{m}<x_n<\frac{1}{m}.\]
Thus \[1-\varepsilon<a^{-1 / m}<a^{x_n}<a^{1 / m}<1+\varepsilon\] which proves \(|a^{x_n}-1|<\varepsilon\) for any \(n \geq N_{m,\varepsilon}\) proving that \[\lim_{n \to \infty}a^{x_n}=1.\]
If \(0<a<1\) we note that \[\lim_{n \to \infty}{a^{x_n}}=\lim_{n \to \infty}\frac{1}{\left(\frac{1}{a}\right)^{x_n}}\] and this completes the proof of the proposition. $$\tag*{$\blacksquare$}$$
Let \(a_1,a_2,\ldots,a_n \geq 0\) be given.
Arithmetic mean. We define arithmetic mean of \(a_1,a_2,\ldots,a_n\) by \[A_n=\frac{a_1+a_2+\ldots+a_n}{n}.\]
Geometric mean. We define geometric mean of \(a_1,a_2,\ldots,a_n\) by \[G_n=\sqrt[n]{a_1\cdot a_2 \cdot \ldots \cdot a_n}.\]
Theorem. For any \(n \in \mathbb{N}\) we have \[G_n \leq A_n.\]
Proof. For \(n=2\) observe that \[(a-b)^2 \geq 0,\]
since \[a^2-2ab+b^2 \geq 0 \ \ \iff \ \ ab \leq \frac{a^2+b^2}{2}.\]
Taking \(a=\sqrt{a_1}\) and \(b=\sqrt{a_2}\) we obtain \[A_2=\frac{a_1+a_2}{2}=\frac{(\sqrt{a_1})^2+(\sqrt{a_2})^2}{2} \geq \sqrt{a_1a_2}=G_2.\]
Case \(n=4\). Note that \[\begin{aligned} &A_4=\frac{a_1+a_2+a_3+a_4}{4}=\frac{1}{2}\left(\frac{a_1+a_2}{2}+\frac{a_3+a_4}{2}\right) \\& \underbrace{\geq}_{{\color{red}A_2 \geq G_2}}\left(\left(a_1a_2\right)^{1/2}\left(a_3a_4\right)^{1/2}\right)^{1 / 2}=(a_1a_2a_3a_4)^{1 / 4}=G_4. \end{aligned}\]
Case \(n=8\). Let us use \(A_4 \geq G_4\) and \(A_2 \geq G_2\) to prove \(A_8 \geq G_8\). \[\begin{aligned} &A_8=\frac{a_1+\ldots+a_8}{8}=\frac{1}{2}\left(\frac{a_1+\ldots+a_4}{4}+\frac{a_5+\ldots+a_8}{4}\right) \\& \underbrace{\geq}_{{\color{red}A_2 \geq G_2}}\left(\frac{a_1+\ldots+a_4}{4}\frac{a_5+\ldots+a_8}{4}\right)^{1 / 2} \\& \underbrace{\geq}_{{\color{red}A_4 \geq G_4}}\left(\left(a_1\ldots a_4\right)^{1 / 4}\left(a_5 \ldots a_8\right)^{1 / 4}\right)^{1 / 2}=\left(a_1\ldots a_8\right)^{1 / 8}=G_8. \end{aligned}\]
We first use induction to prove \[A_{2^n} \geq G_{2^n}\] for all \(n\in\mathbb N\).
Base step. For \(n=2\) the inequality is true as \[A_2=\frac{a_1+a_2}{2} \geq (a_1a_2)^{1 / 2}=G_2.\]
Let \(P(n)\) be the statement that \(A_{2^n} \geq G_{2^n}\) holds for some \(n\in\mathbb N\).
Inductive step. Now we prove that \({\color{red}P(n) \Longrightarrow P(n+1)}\). Indeed, \[\begin{aligned} A_{2^{n+1}}=&\frac{a_1+\ldots+a_{2^{n+1}}}{2^{n+1}}=\frac{1}{2}\left(\frac{a_1+\ldots+a_{2^n}}{2^n}+\frac{a_{2^n+1}+\ldots+a_{2^{n+1}}}{2^n}\right) \\& \underbrace{\geq}_{{\color{red}A_2 \geq G_2}}\left(\frac{a_1+\ldots+a_{2^n}}{2^n}\frac{a_{2^n+1}+\ldots+a_{2^{n+1}}}{2^n}\right)^{1 / 2} \\& \underbrace{\geq}_{{\color{red}A_{2^n} \geq G_{2^n}}}\left(\left(a_1\ldots a_{2^n}\right)^{1 / 2^{n}}\left(a_{2^n+1} \ldots a_{2^{n+1}}\right)^{1 / 2^n}\right)^{1 / 2}\\ &=\left(a_1\ldots a_{2^{n+1}}\right)^{1 / 2^{n+1}}=G_{2^{n+1}}. \end{aligned}\]
Now we have to show that
\[A_n \geq G_n\]. for all \(n\in\mathbb N\).
We first observe that the following downwards induction holds.
Let \(Q(n)\) be the statement that \[A_n \geq G_n\] holds for some \(n \in \mathbb{N}\). Then \[Q(n-1)\] is also true.
This will follow from the so-called bootstrap phenomenon.
Note that (by \(A_n \geq G_n\) with \(a_1,a_2,\ldots, a_{n-1}, a_n=A_{n-1}\)) one has \[\frac{a_1+\ldots+a_{n-1}+A_{n-1}}{n} \geq \left(a_1 \cdot \ldots \cdot a_{n-1}\cdot A_{n-1}\right)^{1 / n}.\]
But \[\frac{a_1+\ldots+a_{n-1}+A_{n-1}}{n}=\frac{(n-1)A_{n-1}+A_{n-1}}{n}=A_{n-1}.\]
Thus we have shown
Bootstrapping inequality. \[A_{n-1} \geq \left(a_1 \cdot \ldots \cdot a_{n-1}\right)^{1 / n}A_{n-1}^{1 / n}.\]
Hence \[A_{n-1}^{1-1 / n} \geq \left(a_1 \cdot \ldots \cdot a_{n-1}\right)^{1 / n}=G_{n-1}^{(n-1)/n},\] thus \(A_{n-1} \geq G_{n-1}\), which means that \(Q(n-1)\) holds.
Now we can show that
Claim ( \(\star\) ). \[A_n \geq G_n \quad \text{ for all } \quad n \in \mathbb{N}.\]
Proof. We know:
\(A^{2^m} \geq G_{2^m}\) for all \(m \in \mathbb{N}\),
if \(A_k \geq G_k\) holds for some \(k \in \mathbb{N}\), then also holds for \(k-1\), i.e. \(A_{k-1} \geq G_{k-1}\) is true.
Concluding, we can easily prove Claim (\(\star\)). Fix \(n \in \mathbb{N}\) and choose the smallest \(m \in \mathbb{N}\) so that \[2^{m-1}<n \leq 2^{m}.\]
By (1) we know \(A_{2^m} \geq G_{2^m}\) holds.
Claim ( \(\star\) ). \[A_n \geq G_n \quad \text{ for all } \quad n \in \mathbb{N}.\]
By (2) with \(k=2^m\) we deduce \[\begin{aligned} A_{2^m} \geq G_{2^m} \quad \text{ implies } \quad A_{2^m-1} \geq G_{2^m-1}. \end{aligned}\] Repeating \[\begin{aligned} A_{2^m-1} \geq G_{2^m-1} \quad \text{ implies } \quad A_{2^m-2} \geq G_{2^m-2}. \end{aligned}\]
We now apply (2) as many times until we reach \(A_{n} \geq G_{n}\) and the proof is finally completed. $$\tag*{$\blacksquare$}$$
Let \(a_1,a_2,\ldots,a_n > 0\) be given. We have the following means.
Arithmetic mean. \[A_n=\frac{a_1+\ldots+a_n}{n};\]
Geometric mean. \[G_n=\left(a_1 \cdot \ldots \cdot a_{n}\right)^{1 / n};\]
Harmonic mean. \[H_n=\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n}};\]
Quadratic mean. \[Q_n=\left(\frac{a_1^2+\ldots+a_n^2}{n}\right)^{1 / 2}.\]
Theorem. For all \(n \in \mathbb{N}\) and \(a_1,a_2,\ldots,a_n > 0\) we have \[\min(a_1,\ldots,a_n) \leq H_n \leq G_n \leq A_n \leq Q_n \leq \max(a_1,\ldots,a_n).\]
Proof. We will proceed in several steps.
We have proved that \(A_n \geq G_n\).
To prove \(H_n \leq G_n\) we apply \(A_n \geq G_n\) with \[\frac{1}{a_1},\frac{1}{a_2},\ldots,\frac{1}{a_n}.\] We obtain \[\begin{aligned} G_n^{-1}=\left(\frac{1}{a_1}\cdot \frac{1}{a_2}\cdot \ldots \cdot \frac{1}{a_n}\right)^{1 / n} \leq \frac{\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n}}{n}=H_n^{-1}, \end{aligned}\] thus \(H_n \leq G_n\).
To prove inequality \(A_n \leq Q_n\) consider the relation \[\begin{aligned} (a_1+a_2+\ldots+a_n)^2&=a_1^2+a_2^2+\ldots+a_n^2\\ &+2(a_1a_1+a_1a_3+\ldots+a_1a_n)\\ &+2(a_2a_3+a_2a_4+\ldots+a_2a_n)\\ & +\ldots+ 2(a_{n-2}a_{n-1}+a_{n-2}a_n)+2a_{n-1}a_n. \end{aligned}\]
Since \(2a_ia_j \leq a_i^2+a_j^2\), thus \[\begin{aligned} (a_1+\ldots+a_n)^2 \leq n(a_1^2+\ldots+a_n^2). \end{aligned}\]
Hence \[\begin{aligned} a_1+\ldots+a_n \leq \left(n(a_1^2+\ldots+a_n^2)\right)^{1/ 2}, \end{aligned}\]
and consequently \(A_n \leq Q_n\).
Finally wlog suppose that \[0<a_1 \leq a_2 \leq \ldots \leq a_n.\]
then \(a_1=\min(a_1,\ldots,a_n)\), and \(a_n=\max(a_1,\ldots,a_n)\). Hence \[\begin{aligned} H_n=\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n}} \geq n\frac{a_1}{n}=a_1, \end{aligned}\] and \[\begin{aligned} Q_n=\left(\frac{a_1^2+\ldots+a_n^2}{n}\right)^{1/2} \leq \left(\frac{na_n^2}{n}\right)^{1/2}=a_n. \end{aligned}\] The proof is completed. $$\tag*{$\blacksquare$}$$
Example. Prove that for any \(x,y,z>0\) we have \[\frac{x^2}{yz}+\frac{y^2}{xz}+\frac{z^2}{xy} \geq 3.\]
Solution. Consider the numbers \(\frac{x^2}{yz}\), \(\frac{y^2}{xz}\), \(\frac{z^2}{xy}\). Then \[A_3=\frac{\frac{x^2}{yz}+\frac{y^2}{xz}+\frac{z^2}{xy}}{3},\] \[G_3=\sqrt[3]{\frac{x^2}{yz}\cdot\frac{y^2}{xz}\cdot\frac{z^2}{xy}}=1,\] so our inequality is a consequence of \(A_3 \geq G_3\).$$\tag*{$\blacksquare$}$$
Example. If the product of \(n\) positive real numbers is \(1\), then their sum is at least \(n\).
Solution. Let \(a_1,\ldots,a_n>0\) be the numbers such that \[G_n=\sqrt[n]{a_1\cdots a_n}=1,\] so by \(A_n \geq G_n\), \[a_1+\ldots+a_n \geq n G_n=n.\] $$\tag*{$\blacksquare$}$$
Bernoulli inequality. If \(x>-1\) and \(n \in \mathbb{N}\), then one has \[(1+x)^n \geq 1+nx.\]
Proof. We will use \(A_n \geq G_n\) with \[a_1=a_2=\ldots=a_{n-1}=1 \quad \text{ and }\quad a_n=1+nx.\] Indeed, \[\begin{aligned} A_n=\frac{a_1+\ldots+a_n}{n}=\frac{\overbrace{1+\ldots+1}^{n-1 \text{ times}}+1+nx}{n}=\frac{n(1 +x)}{n}=1+x. \end{aligned}\]
On the other hand \[\begin{aligned} (1+x)=A_n \geq G_n=\left(\overbrace{1 \cdot 1 \cdot \ldots \cdot 1}^{n-1 \text{ times}} \cdot (1+nx)\right)^{1 / n}=(1+nx)^{1 / n}, \end{aligned}\]
which implies \[\begin{aligned} (1+x)^n \geq 1+nx, \end{aligned}\] and we are done. $$\tag*{$\blacksquare$}$$
Our aim will be to built tools and generalize Bernoulli’s inequality. We show that the following is true.
Bernoulli inequality - generalization. If \(-1<x \neq 0\) and \(a>1\) or \(a<0\), then \[(1+x)^a>1+ax.\]
If \(-1<x \neq 0\) and \(0 <a<1\), then \[(1+x)^a<1+ax.\]
Cauchy–Schwarz inequality. For any real numbers \(a_1,\ldots,a_n\) and \(b_1,\ldots,b_n\) one has \[\begin{aligned} \left(\sum_{j=1}^na_jb_j\right)^2 \leq \left(\sum_{j=1}^na_j^2\right)\left(\sum_{j=1}^nb_j^2\right). \end{aligned}\]
Proof. Consider the polynomial \[\begin{aligned} &0 \leq (a_1x+b_1)^2+(a_2x+b_2)^2+\ldots+(a_nx+b_n)^2=\\& (a_1^2+\ldots+a_n^2)x^2+2(a_1b_1+a_2b_2+\ldots+a_nb_n)x+(b_1^2+\ldots+b_n^2). \end{aligned}\]
Since the polynomial is nonnegative \[\begin{aligned} \Delta=4(a_1b_1+\ldots+a_nb_n)^2-4(a_1^2+\ldots+a_n^2)(b_1^2+\ldots+b_n^2) \leq 0 \end{aligned}\] and we are done. $$\tag*{$\blacksquare$}$$
Minkowski’s inequality. For any real numbers \(a_1,\ldots,a_n\) and \(b_1,\ldots,b_n\) one has \[\left(\sum_{j=1}^n|a_j+b_j|^2\right)^{1 / 2} \leq \left(\sum_{j=1}^n|a_j|^2\right)^{1 / 2}+\left(\sum_{j=1}^n|b_j|^2\right)^{1 / 2}.\]
Proof. Let \[S_n=\left(\sum_{j=1}^{n}|a_j+b_j|^2\right)^{1 / 2}.\]
Then \[\begin{aligned} S_n^2&=\sum_{j=1}^{n}|a_j+b_j|^2=\sum_{j=1}^{n}|a_j+b_j||a_j+b_j| \\&\leq \sum_{j=1}^{n}|a_j+b_j||a_j|+\sum_{j=1}^{n}|a_j+b_j||b_j|. \end{aligned}\]
By the Cauchy–Schwarz inequality, \[\begin{aligned} S_n^2&\leq \\ \underbrace{\left(\sum_{j=1}^n |a_j+b_j|^2\right)^{1 / 2}}_{=S_n}\left(\sum_{j=1}^n |a_j|^2\right)^{1 / 2} &+\underbrace{\left(\sum_{j=1}^n |a_j+b_j|^2\right)^{1 / 2}}_{=S_n}\left(\sum_{j=1}^n |b_j|^2\right)^{1 / 2} \end{aligned}\]
\[\begin{aligned} =S_n \left(\left(\sum_{j=1}^n |a_j|^2\right)^{1 / 2}+\left(\sum_{j=1}^n |b_j|^2\right)^{1 / 2}\right). \end{aligned}\]
Thus we have proved a bootstrap inequality, i.e. \[S_n^2 \leq S_n \left(\left(\sum_{j=1}^n |a_j|^2\right)^{1 / 2}+\left(\sum_{j=1}^n |b_j|^2\right)^{1 / 2}\right).\]
Hence \[S_n \leq \left(\sum_{j=1}^n |a_j|^2\right)^{1 / 2}+\left(\sum_{j=1}^n |b_j|^2\right)^{1 / 2}.\] $$\tag*{$\blacksquare$}$$
Theorem. For all positive real numbers \(a_1,a_2,\ldots,a_n\) and all positive weights \(q_1,q_2,\ldots,q_n\) satisfying the following convexity condition \[{\color{red}q_1+\ldots+q_n=1,}\]
we have \[{\color{blue}a_1^{q_1}\cdot\ldots\cdot a_n^{q_n} \leq q_1a_1+\ldots+q_na_n.}\]
If \(q_1=q_2=\ldots=q_n=\frac{1}{n}\), then we have \[a_1^{q_1}\cdot\ldots\cdot a_n^{q_n}=(a_1\cdot\ldots\cdot a_n)^{1/n}\le \frac{a_1+\ldots+ a_n}{n}=q_1a_1+\ldots+q_na_n,\] which recovers the inequality between geometric and arithmetic means.
Proof: We first assume \[q_1,\ldots,q_n \in \mathbb{Q} \quad \text{ and } \quad q_1,\ldots,q_n>0.\]
We can assume that \({\color{red}q_i=\frac{k_i}{m}}\) for \(1 \leq i \leq n\) and \[{\color{red}k_1+\ldots+k_n=m.}\]
Invoking the inequality between geometric and arithmetic means we obtain \[\begin{aligned} \sum_{i=1}^{n}q_ia_i&=k_1\frac{a_1}{m}+\ldots+k_n\frac{a_n}{m} \\ &\geq m \left(\left(\frac{a_1}{m}\right)^{k_1}\cdot \ldots \cdot \left(\frac{a_n}{m}\right)^{k_n}\right)^{1 / m}\\ & = a_1^{k_1 / m}\cdot \ldots \cdot a_n^{k_n / m}\\ &=a_1^{q_1}\cdot \ldots \cdot a_n^{q_n}. \end{aligned}\]
If now all weights \(q_1,\ldots,q_n>0\) are real numbers, then for any \(1 \leq i \leq n\), we choose a sequence of positive rationals \((q_{i,k})_{k \in \mathbb{N}}\) so that \[\lim_{k \to \infty}q_{i,k}=q_i\] and so that \[{\color{red}\sum_{i=1}^{n}q_{i,k}=1} \quad \text{ for all }\quad k \in \mathbb{N}.\]
Then by the previous part \[a_1^{q_{i,k}}\cdot \ldots \cdot a_n^{q_{n,k}} \leq q_{1,k}a_1+\ldots+q_{n,k}a_{n}.\]
Passing with \(k \to \infty\) we conclude that \[{\color{blue}a_1^{q_1}\cdot\ldots\cdot a_n^{q_n} \leq q_1a_1+\ldots+q_na_n.}\] Here we used the following fact: if \(a>0\) and \(\lim_{n \to \infty}x_n=x_0\), then \(\lim_{n \to \infty}a^{x_n}=a^{x_0}\).$$\tag*{$\blacksquare$}$$
Hölder’s inequality. Let \(1<p,q<\infty\) be such that \(\frac{1}{p}+\frac{1}{q}=1\). Then for any real numbers \(x_1,\ldots,x_n\) and \(y_1,\ldots,y_n\) one has \[\begin{aligned} \sum_{j=1}^n|x_jy_j| \leq \Big(\sum_{j=1}^n|x_j|^p\Big)^{1 / p}\Big(\sum_{j=1}^n|y_j|^q\Big)^{1 / q}. \end{aligned}\]
Proof. By the previous theorem for any \(a_1,b_1>0\) we have \[a_1^{\frac 1 p}b_1^{\frac 1 q} \leq \frac{1}{p}a_1+\frac{1}{q}b_1,\] which for \(a_1=a^p\) and \(b_1=b^q\) yields
(*). \[ab \leq \frac{1}{p}a^p+\frac{1}{q}b^q.\]
Let
\[a_j:=\frac{|x_j|}{\big(\sum_{j=1}^{n}|x_j|^{p}\big)^{1 / p}}, \ \ \ b_j:=\frac{|y_j|}{\big(\sum_{j=1}^{n}|y_j|^{q}\big)^{1 / q}}\].
Applying (*) to each \(1 \leq j \leq n\) one gets \[\begin{aligned} \sum_{j=1}^{n}a_jb_j &\leq \sum_{j=1}^{n}\left(\frac{1}{p}a_j^p+\frac{1}{q}b_j^{q}\right)\\& =\sum_{j=1}^{n}\left(\frac{|x_j|^p}{p\big(\sum_{j=1}^{n}|x_j|^{p}\big)}+\frac{|y_j^q|}{q\big(\sum_{j=1}^{n}|y_j|^{q}\big)}\right)\\&=\frac{1}{p}\frac{\sum_{j=1}^{n}|x_j|^p}{\sum_{j=1}^{n}|x_j|^p}+\frac{1}{q}\frac{\sum_{j=1}^{n}|y_j|^q}{\sum_{j=1}^{n}|y_j|^q}=\frac{1}{p}+\frac{1}{q}= 1. \end{aligned}\]
Thus we have proved \[\sum_{j=1}^{n}a_jb_j \leq 1,\]
which is equivalent to
\[\sum_{j=1}^n|x_jy_j| \leq \Big(\sum_{j=1}^n|x_j|^p\Big)^{1 / p}\Big(\sum_{j=1}^n|y_j|^q\Big)^{1 / q}\] and the proof of Hölder’s inequality is completed. $$\tag*{$\blacksquare$}$$