Inequality Identities (mean, norm, sum, integral)

$$\frac{n}{\underset{k=1}{\overset{n}{\sum}}\frac{1}{a_k}} \leq$$

$$\sqrt[n]{\underset{k=1}{\overset{n}{\prod}}a_k}$$

$$\sqrt[n]{\underset{k=1}{\overset{n}{\prod}}a_k} \leq$$

$$\frac{1}{n}\underset{k=1}{\overset{n}{\sum}}a_k$$

$$\frac{1}{n}\underset{k=1}{\overset{n}{\sum}}a_k \leq$$

$$\sqrt{\frac{1}{n}\underset{k=1}{\overset{n}{\sum}}a_k^2}$$

p < q \land p,q \neq 0 \land \forall k \in \mathcal{I}_k \space (a_k > 0) \implies \left( \frac{1}{n} \underset{k=1}{\overset{n}{\sum}} a_k^p \right)^{\frac{1}{p}} \leq

$$\left( \frac{1}{n} \underset{k=1}{\overset{n}{\sum}} a_k^q \right)^{\frac{1}{q}}$$

$$\left|a-b\right| \geq$$

$$\left|\left|a\right|-\left|b\right|\right|$$

$$p \geq 1 \implies \left( \underset{k=1}{\overset{n}{\sum}} \left|a_k+b_k\right|^p \right)^{\frac{1}{p}} \leq$$

$$\left( \underset{k=1}{\overset{n}{\sum}} \left| a_k \right|^p \right)^{\frac{1}{p}} + \left( \underset{k=1}{\overset{n}{\sum}} \left| b_k \right|^p \right)^{\frac{1}{p}}$$

$$p,q \geq 1 \land \frac{1}{p} + \frac{1}{q} = 1 \implies \underset{k=1}{\overset{n}{\sum}} \left| a_kb_k \right| \leq$$

$$\left( \underset{k=1}{\overset{n}{\sum}} \left| a_k \right|^p \right)^{\frac{1}{p}} \left( \underset{k=1}{\overset{n}{\sum}} \left| b_k \right|^q \right)^{\frac{1}{q}}$$

$$\left(\underset{k=1}{\overset{n}{\sum}}a_kb_k\right)^2 \leq$$

$$\left(\underset{k=1}{\overset{n}{\sum}}a_k^2\right)\left(\underset{k=1}{\overset{n}{\sum}}b_k^2\right)$$

$$\left( \int f(x)g(x) \space dx \right)^2 \leq$$

$$\int f(x)^2 dx \cdot \int g(x)^2 dx$$