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Question Number 71508 by Cmr 237 last updated on 16/Oct/19

$$\mathrm{montrer}\mathrm{que}:\mathrm{\forall }\mathrm{a},\mathrm{b}\in \mathbb{R}\mathrm{ona}$$$$2\mid \mathrm{ab}\mid ⩽{\mathrm{a}}^2+{\mathrm{b}}^2$$$$\mathrm{Endeduire}\mathrm{que}\mathrm{\forall }{\mathrm{x}}_1,...,{\mathrm{x}}_{\mathrm{n}}\in \mathbb{R}\mathrm{on}\mathrm{a}:$$$${(\underset{\mathrm{i}=1}{\sum ^{\mathrm{n}}}\mid {\mathrm{x}}_{\mathrm{i}}\mid )}^2⩽\mathrm{n}\underset{\mathrm{i}=1}{\sum ^{\mathrm{n}}}{\mathrm{x}}_{\mathrm{i}}^2$$$$\mathbf{please}\mathbf{i}\mathbf{need}\mathbf{help}$$

Commented by Cmr 237 last updated on 16/Oct/19

$$\mathrm{please}\mathrm{can}\mathrm{you}\mathrm{prove}\mathrm{it}?$$

Commented by Prithwish sen last updated on 16/Oct/19

$$\mathbf{let}$$$$\mathbf{k}=\frac{{\mathbf{x}}_1+{\mathbf{x}}_2+.......{\mathbf{x}}_{\mathbf{n}}}{\mathbf{n}}.....\left(\mathbf{i}\right)$$$$\mathrm{now}\frac{{\mathrm{x}}_1}{\mathrm{k}},\frac{{\mathrm{x}}_2}{\mathrm{k}},........\frac{{\mathrm{x}}_{\mathrm{n}}}{\mathrm{k}}\mathrm{all}\mathrm{are}\mathrm{positive}\mathrm{and}\mathrm{not}\mathrm{all}$$$$\mathrm{of}\mathrm{them}\mathrm{equal}\mathrm{to}1$$$${\left[\frac{{\mathbf{x}}_{\mathbf{i}}}{\mathbf{k}}\right]}^2-1⩾0\mathrm{equality}\mathrm{holds}\mathrm{only}\frac{{\mathrm{x}}_{\mathrm{i}}}{\mathrm{k}}=1(\mathrm{i}\in \mathbb{N})$$$$\mathrm{now}{\left[\frac{{\mathrm{x}}_1}{\mathrm{k}}\right]}^2+{\left[\frac{{\mathrm{x}}_2}{\mathrm{k}}\right]}^2+.....{\left[\frac{{\mathrm{x}}_{\mathrm{n}}}{\mathrm{k}}\right]}^2-\mathrm{n}⩾2[\frac{{\mathrm{x}}_1}{\mathrm{k}}+\frac{{\mathrm{x}}_2}{\mathrm{k}}+.....+\frac{{\mathrm{x}}_{\mathrm{n}}}{\mathrm{k}}-\mathrm{n}]\left(\mathbf{from}\mathbf{given}\mathbf{condition}\right)$$$$\mathrm{or},\frac{{\mathrm{x}}_1^2+{\mathrm{x}}_2^2+....+{\mathrm{x}}_{\mathrm{n}}^2}{{\mathrm{k}}^2}-\mathrm{n}⩾2(\mathrm{n}-\mathrm{n})\mathbf{from}\left(\mathbf{i}\right)$$$$\mathbf{or}\frac{{\mathbf{x}}_1^2+{\mathbf{x}}_2^2+.......+{\mathbf{x}}_{\mathbf{n}}^2}{\mathbf{n}}⩾{\mathbf{k}}^2$$$$\Rightarrow {\mathbf{x}}_1^2+{\mathbf{x}}_2^2+......+{\mathbf{x}}_{\mathbf{n}}^2⩾\mathrm{n}{\left(\frac{{\mathrm{x}}_1+{\mathrm{x}}_2+.....+{\mathrm{x}}_{\mathrm{n}}}{\mathrm{n}}\right)}^2$$$$\text{Double subscripts: use braces to clarify}$$

Answered by turbo msup by abdo last updated on 17/Oct/19

$$wehsve\sum _{i=1}^n{a}_i{b}_i⩽{\left(\sum _{i=1}^n{a}_i^2\right)}^{\frac{1}{2}}\times {\left(\sum _{i=1}^n{b}_i^2\right)}^{\frac{1}{2}}\left(holder\right)$$$$forallnumberspositifs\left(a_i\right)snd\left(b_i\right)$$$$let{a}_i=1snd{b}_i=\mid x_i\mid \Rightarrow$$$$\sum _{i=1}^n\mid x_i\mid ⩽\sqrt{n}\times {\left(\sum _{i=1}^n{x}_i^2\right)}^{\frac{1}{2}}\Rightarrow$$$${(\sum _{i=1}^n\mid x_i\mid )}^2⩽n\times \left(\sum _{i=1}^n{x}_i^2\right)$$$$$$

Commented by Prithwish sen last updated on 17/Oct/19

$$\mathrm{Thank}\mathrm{you}\mathrm{sir}.$$

Commented by mathmax by abdo last updated on 18/Oct/19

$$youarewelcome.$$

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