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Question Number 71508 by Cmr 237 last updated on 16/Oct/19
montrer que:∀a,b∈R ona 2∣ab∣≤a^2 +b^2 Endeduire que ∀x_1 ,...,x_n ∈R on a: (Σ_(i=1) ^n ∣x_i ∣)^2 ≤nΣ_(i=1) ^n x_i ^2 please i need help
$$\mathrm{montrer}\:\mathrm{que}:\forall\mathrm{a},\mathrm{b}\in\mathbb{R}\:\mathrm{ona} \\ $$$$\mathrm{2}\mid\mathrm{ab}\mid\leqslant\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \\ $$$$\mathrm{Endeduire}\:\mathrm{que}\:\forall\mathrm{x}_{\mathrm{1}} ,…,\mathrm{x}_{\mathrm{n}} \in\mathbb{R}\:\mathrm{on}\:\mathrm{a}: \\ $$$$\left(\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mid\mathrm{x}_{\mathrm{i}} \mid\right)^{\mathrm{2}} \leqslant\mathrm{n}\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{x}_{\mathrm{i}} ^{\mathrm{2}} \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{need}}\:\boldsymbol{\mathrm{help}} \\ $$
Commented by Cmr 237 last updated on 16/Oct/19
please can you prove it?
$$\mathrm{please}\:\mathrm{can}\:\mathrm{you}\:\mathrm{prove}\:\mathrm{it}? \\ $$
Commented by Prithwish sen last updated on 16/Oct/19
let k= ((x_1 +x_2 +.......x_n )/n) .....(i) now (x_1 /k),(x_2 /k),........(x_n /k) all are positive and not all of them equal to 1 [(x_i /k)]^2 −1≥0 equality holds only (x_i /k) = 1 (i∈N) now [(x_1 /k)]^2 +[(x_2 /k)]^2 +.....[(x_n /k)]^2 −n ≥ 2[(x_1 /k)+(x_2 /k)+.....+(x_n /k)−n] (from given condition) or, ((x_1 ^2 +x_2 ^2 +.... +x_n ^2 )/k^2 ) −n ≥ 2(n−n) from (i) or ((x_1 ^2 +x_2 ^2 +.......+x_n ^2 )/n) ≥ k^2 ⇒ x_1 ^2 +x_2 ^2 +......+x_n ^2 ≥ n(((x_1 +x_2 +.....+x_n )/n))^2 ⇒n𝚺_1 ^n (x_i )^2 ≥ (𝚺_1 ^n x_i )^2 Hence proved.
$$\boldsymbol{\mathrm{let}} \\ $$$$\boldsymbol{\mathrm{k}}=\:\frac{\boldsymbol{\mathrm{x}}_{\mathrm{1}} +\boldsymbol{\mathrm{x}}_{\mathrm{2}} +…….\boldsymbol{\mathrm{x}}_{\boldsymbol{\mathrm{n}}} }{\boldsymbol{\mathrm{n}}}\:…..\left(\boldsymbol{\mathrm{i}}\right) \\ $$$$\mathrm{now}\:\frac{\mathrm{x}_{\mathrm{1}} }{\mathrm{k}},\frac{\mathrm{x}_{\mathrm{2}} }{\mathrm{k}},……..\frac{\mathrm{x}_{\mathrm{n}} }{\mathrm{k}}\:\mathrm{all}\:\mathrm{are}\:\mathrm{positive}\:\mathrm{and}\:\mathrm{not}\:\mathrm{all} \\ $$$$\mathrm{of}\:\mathrm{them}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{1} \\ $$$$\left[\frac{\boldsymbol{\mathrm{x}}_{\boldsymbol{\mathrm{i}}} }{\boldsymbol{\mathrm{k}}}\right]^{\mathrm{2}} −\mathrm{1}\geqslant\mathrm{0}\:\:\mathrm{equality}\:\mathrm{holds}\:\mathrm{only}\:\frac{\mathrm{x}_{\mathrm{i}} }{\mathrm{k}}\:=\:\mathrm{1}\:\:\left(\mathrm{i}\in\mathbb{N}\right)\: \\ $$$$\mathrm{now}\:\left[\frac{\mathrm{x}_{\mathrm{1}} }{\mathrm{k}}\right]^{\mathrm{2}} +\left[\frac{\mathrm{x}_{\mathrm{2}} }{\mathrm{k}}\right]^{\mathrm{2}} +…..\left[\frac{\mathrm{x}_{\mathrm{n}} }{\mathrm{k}}\right]^{\mathrm{2}} −\mathrm{n}\:\geqslant\:\mathrm{2}\left[\frac{\mathrm{x}_{\mathrm{1}} }{\mathrm{k}}+\frac{\mathrm{x}_{\mathrm{2}} }{\mathrm{k}}+…..+\frac{\mathrm{x}_{\mathrm{n}} }{\mathrm{k}}−\mathrm{n}\right]\:\:\left(\boldsymbol{\mathrm{from}}\:\boldsymbol{\mathrm{given}}\:\boldsymbol{\mathrm{condition}}\right) \\ $$$$\mathrm{or},\:\frac{\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{x}_{\mathrm{2}} ^{\mathrm{2}} +….\:+\mathrm{x}_{\mathrm{n}} ^{\mathrm{2}} }{\mathrm{k}^{\mathrm{2}} }\:−\mathrm{n}\:\geqslant\:\mathrm{2}\left(\mathrm{n}−\mathrm{n}\right)\:\:\boldsymbol{\mathrm{from}}\:\left(\boldsymbol{\mathrm{i}}\right) \\ $$$$\boldsymbol{\mathrm{or}}\:\frac{\boldsymbol{\mathrm{x}}_{\mathrm{1}} ^{\mathrm{2}} +\boldsymbol{\mathrm{x}}_{\mathrm{2}} ^{\mathrm{2}} +…….+\boldsymbol{\mathrm{x}}_{\boldsymbol{\mathrm{n}}} ^{\mathrm{2}} }{\boldsymbol{\mathrm{n}}}\:\geqslant\:\boldsymbol{\mathrm{k}}^{\mathrm{2}} \\ $$$$\Rightarrow\:\boldsymbol{\mathrm{x}}_{\mathrm{1}} ^{\mathrm{2}} +\boldsymbol{\mathrm{x}}_{\mathrm{2}} ^{\mathrm{2}} +……+\boldsymbol{\mathrm{x}}_{\boldsymbol{\mathrm{n}}} ^{\mathrm{2}} \:\geqslant\:\mathrm{n}\left(\frac{\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} +…..+\mathrm{x}_{\mathrm{n}} }{\mathrm{n}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\boldsymbol{\mathrm{n}}\underset{\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\boldsymbol{\sum}}}\left(\boldsymbol{\mathrm{x}}_{\boldsymbol{\mathrm{i}}} \right)^{\mathrm{2}} \geqslant\:\left(\underset{\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\boldsymbol{\sum}}\mathrm{x}}_{\boldsymbol{\mathrm{i}}} \:\right)^{\mathrm{2}} \:\:\boldsymbol{\mathrm{Hence}}\:\boldsymbol{\mathrm{proved}}. \\ $$
Answered by turbo msup by abdo last updated on 17/Oct/19
we hsve Σ_(i=1) ^n a_i b_i ≤(Σ_(i=1) ^n a_i ^2 )^(1/2) ×(Σ_(i=1) ^n b_i ^2 )^(1/2) (holder) for all numbers positifs (a_i )snd (b_i ) let a_i =1 snd b_i =∣x_i ∣ ⇒ Σ_(i=1) ^n ∣x_i ∣≤(√n)×(Σ_(i=1) ^n x_i ^2 )^(1/2) ⇒ (Σ_(i=1) ^n ∣x_i ∣)^2 ≤n ×(Σ_(i=1) ^n x_i ^2 )
$${we}\:{hsve}\:\sum_{{i}=\mathrm{1}} ^{{n}} {a}_{{i}} {b}_{{i}} \leqslant\left(\sum_{{i}=\mathrm{1}} ^{{n}} {a}_{{i}} ^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} ×\left(\sum_{{i}=\mathrm{1}} ^{{n}} {b}_{{i}} ^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left({holder}\right) \\ $$$${for}\:{all}\:{numbers}\:{positifs}\:\left({a}_{{i}} \right){snd}\:\left({b}_{{i}} \right) \\ $$$${let}\:{a}_{{i}} =\mathrm{1}\:{snd}\:{b}_{{i}} =\mid{x}_{{i}} \mid\:\Rightarrow \\ $$$$\sum_{{i}=\mathrm{1}} ^{{n}} \mid{x}_{{i}} \mid\leqslant\sqrt{{n}}×\left(\sum_{{i}=\mathrm{1}} ^{{n}} {x}_{{i}} ^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:\Rightarrow \\ $$$$\left(\sum_{{i}=\mathrm{1}} ^{{n}} \mid{x}_{{i}} \mid\right)^{\mathrm{2}} \leqslant{n}\:×\left(\sum_{{i}=\mathrm{1}} ^{{n}} \:{x}_{{i}} ^{\mathrm{2}} \right) \\ $$$$ \\ $$
Commented by Prithwish sen last updated on 17/Oct/19
Thank you sir.
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by mathmax by abdo last updated on 18/Oct/19
you are welcome.
$${you}\:{are}\:{welcome}. \\ $$

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