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Question Number 4719 by prakash jain last updated on 29/Feb/16
Prove that ((a_1 +a_2 +...+a_n )/n)≤(√((a_1 ^2 +a_2 ^2 +...+a_n ^2 )/n)) with equality holding iff a_1 =a_2 =...=a_n .
$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\frac{{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +…+{a}_{{n}} }{{n}}\leqslant\sqrt{\frac{{a}_{\mathrm{1}} ^{\mathrm{2}} +{a}_{\mathrm{2}} ^{\mathrm{2}} +…+{a}_{{n}} ^{\mathrm{2}} }{{n}}} \\ $$$$\mathrm{with}\:\mathrm{equality}\:\mathrm{holding}\:\mathrm{iff}\:{a}_{\mathrm{1}} ={a}_{\mathrm{2}} =…={a}_{{n}} . \\ $$
Commented by Yozzii last updated on 29/Feb/16
u=(a_1 ,a_2 ,a_3 ,...,a_n ),m=(1/n)(1,1,1,...,1) ⇒u.m=(1/n)Σ_(r=1) ^n a_r ∣u∣=(√(Σ_(r=1) ^n a_r ^2 )),∣m∣=(1/n)(√(Σ_(i=1) ^n 1))=(1/( (√n))) ∴ since cosα=((u.m)/(∣u∣∣m∣)) and ∣cosα∣≤1 (−π<α≤π) ⇒∣u.m∣≤∣u∣∣m∣ ∴ ∣(1/n)Σ_(r=1) ^n a_r ∣≤(√((Σ_(r=1) ^n a_r ^2 )/n)) This implies that −(√((1/n)(Σ_(r=1) ^n a_r ^2 )))≤(1/n)Σ_(r=1) ^n a_r ≤(√((1/n)(Σ_(r=1) ^n a_r ^2 ))) For all a_r ∈R we then deduce (1/n)Σ_(r=1) ^n a_r ≤(√((1/n)(Σ_(r=1) ^n a_r ^2 ))) If a_1 =a_2 =...=a_n ⇒(1/n)×na_1 ≤(√((1/n)×na_1 ^2 )) a_1 ≤a_1 ⇒equality If equailty occurs ⇒(1/n^2 )(Σa)^2 =(1/n)(Σa^2 ) (Σa)^2 =nΣa^2 From this we obtain cosα=1⇒α=0. ⇒u=m⇒a_1 =a_2 =a_3 =...=a_n =(1/n).
$$\boldsymbol{{u}}=\left({a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,{a}_{\mathrm{3}} ,…,{a}_{{n}} \right),\boldsymbol{{m}}=\frac{\mathrm{1}}{{n}}\left(\mathrm{1},\mathrm{1},\mathrm{1},…,\mathrm{1}\right) \\ $$$$\Rightarrow\boldsymbol{{u}}.\boldsymbol{{m}}=\frac{\mathrm{1}}{{n}}\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{r}} \\ $$$$\mid\boldsymbol{{u}}\mid=\sqrt{\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{r}} ^{\mathrm{2}} },\mid\boldsymbol{{m}}\mid=\frac{\mathrm{1}}{{n}}\sqrt{\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{1}}=\frac{\mathrm{1}}{\:\sqrt{{n}}} \\ $$$$\therefore\:{since}\:{cos}\alpha=\frac{\boldsymbol{{u}}.\boldsymbol{{m}}}{\mid\boldsymbol{{u}}\mid\mid\boldsymbol{{m}}\mid}\:{and}\:\mid{cos}\alpha\mid\leqslant\mathrm{1}\:\left(−\pi<\alpha\leqslant\pi\right) \\ $$$$\Rightarrow\mid\boldsymbol{{u}}.\boldsymbol{{m}}\mid\leqslant\mid\boldsymbol{{u}}\mid\mid\boldsymbol{{m}}\mid \\ $$$$\therefore\:\mid\frac{\mathrm{1}}{{n}}\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{r}} \mid\leqslant\sqrt{\frac{\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{r}} ^{\mathrm{2}} }{{n}}} \\ $$$${This}\:{implies}\:{that}\: \\ $$$$−\sqrt{\frac{\mathrm{1}}{{n}}\left(\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{r}} ^{\mathrm{2}} \right)}\leqslant\frac{\mathrm{1}}{{n}}\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{r}} \leqslant\sqrt{\frac{\mathrm{1}}{{n}}\left(\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{r}} ^{\mathrm{2}} \right)} \\ $$$${For}\:{all}\:{a}_{{r}} \in\mathbb{R}\:{we}\:{then}\:{deduce}\: \\ $$$$\frac{\mathrm{1}}{{n}}\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{r}} \leqslant\sqrt{\frac{\mathrm{1}}{{n}}\left(\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{r}} ^{\mathrm{2}} \right)} \\ $$$$ \\ $$$${If}\:{a}_{\mathrm{1}} ={a}_{\mathrm{2}} =…={a}_{{n}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{n}}×{na}_{\mathrm{1}} \leqslant\sqrt{\frac{\mathrm{1}}{{n}}×{na}_{\mathrm{1}} ^{\mathrm{2}} } \\ $$$${a}_{\mathrm{1}} \leqslant{a}_{\mathrm{1}} \Rightarrow{equality} \\ $$$${If}\:{equailty}\:{occurs} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left(\Sigma{a}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{{n}}\left(\Sigma{a}^{\mathrm{2}} \right) \\ $$$$\left(\Sigma{a}\right)^{\mathrm{2}} ={n}\Sigma{a}^{\mathrm{2}} \\ $$$${From}\:{this}\:{we}\:{obtain}\:{cos}\alpha=\mathrm{1}\Rightarrow\alpha=\mathrm{0}. \\ $$$$\Rightarrow\boldsymbol{{u}}=\boldsymbol{{m}}\Rightarrow{a}_{\mathrm{1}} ={a}_{\mathrm{2}} ={a}_{\mathrm{3}} =…={a}_{{n}} =\frac{\mathrm{1}}{{n}}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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