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Question Number 45353 by pieroo last updated on 12/Oct/18

Prove that p(n)=((a_1 +a_2 +...+a_n )/n) ≥^n (√(a_1 a_2 ...a_n )) ∀ n ∈N

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{p}\left(\mathrm{n}\right)=\frac{\boldsymbol{\mathrm{a}}_{\mathrm{1}} +\boldsymbol{\mathrm{a}}_{\mathrm{2}} +...+\boldsymbol{\mathrm{a}}_{\mathrm{n}} }{\mathrm{n}}\:\geqslant\:^{\mathrm{n}} \sqrt{\boldsymbol{\mathrm{a}}_{\mathrm{1}} \boldsymbol{\mathrm{a}}_{\mathrm{2}} ...\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} } \\ $$$$\forall\:\mathrm{n}\:\in\boldsymbol{\mathrm{N}} \\ $$

Commented by pieroo last updated on 12/Oct/18

please help

$$\mathrm{please}\:\mathrm{help} \\ $$

Commented by Kunal12588 last updated on 12/Oct/18

trying PMI(Induction)

$${trying}\:{PMI}\left({Induction}\right) \\ $$

Commented by pieroo last updated on 14/Oct/18

I still need help urgently please

$$\mathrm{I}\:\mathrm{still}\:\mathrm{need}\:\mathrm{help}\:\mathrm{urgently}\:\mathrm{please} \\ $$

Answered by Kunal12588 last updated on 12/Oct/18

p(n):((a_1 +a_2 +a_3 +...+a_n )/n)≥((a_1 a_2 a_3 ...a_n ))^(1/n) p(1):(a_1 /1)=a_1 (a_1 )^(1/1) =a_1 ∴p(1) satisfies p(n) p(2):((a_1 +a_2 )/2) ((a_1 a_2 ))^(1/2) =(√a_1 )(√a_2 ) (1/2)((√a_1 )−(√a_2 ))^2 ≥0 ⇒ ((a_1 +a_2 −2(√a_1 )(√a_2 ))/2)≥0 ⇒((a_1 +a_2 )/2)≥(√a_1 )(√a_2 ) ∴p(2) satisfies p(n) let us assume p(n) is true for n=k ⇒p(k):((a_1 +a_2 +a_3 +...+a_k )/k)≥((a_1 a_2 a_3 ...a_k ))^(1/k) now we have to show p(k+1):((a_1 +a_2 +a_3 +...+a_(k+1) )/(k+1))≥((a_1 a_2 a_3 ...a_(k+1) ))^(1/(k+1)) please help

$${p}\left({n}\right):\frac{{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} +...+{a}_{{n}} }{{n}}\geqslant\sqrt[{{n}}]{{a}_{\mathrm{1}} {a}_{\mathrm{2}} {a}_{\mathrm{3}} ...{a}_{{n}} } \\ $$$${p}\left(\mathrm{1}\right):\frac{{a}_{\mathrm{1}} }{\mathrm{1}}={a}_{\mathrm{1}} \:\:\:\:\:\:\:\:\:\:\:\sqrt[{\mathrm{1}}]{{a}_{\mathrm{1}} }={a}_{\mathrm{1}} \\ $$$$\therefore{p}\left(\mathrm{1}\right)\:{satisfies}\:{p}\left({n}\right) \\ $$$${p}\left(\mathrm{2}\right):\frac{{a}_{\mathrm{1}} +{a}_{\mathrm{2}} }{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\sqrt[{\mathrm{2}}]{{a}_{\mathrm{1}} {a}_{\mathrm{2}} }=\sqrt{{a}_{\mathrm{1}} }\sqrt{{a}_{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{{a}_{\mathrm{1}} }−\sqrt{{a}_{\mathrm{2}} }\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\Rightarrow\:\frac{{a}_{\mathrm{1}} +{a}_{\mathrm{2}} −\mathrm{2}\sqrt{{a}_{\mathrm{1}} }\sqrt{{a}_{\mathrm{2}} }}{\mathrm{2}}\geqslant\mathrm{0} \\ $$$$\Rightarrow\frac{{a}_{\mathrm{1}} +{a}_{\mathrm{2}} }{\mathrm{2}}\geqslant\sqrt{{a}_{\mathrm{1}} }\sqrt{{a}_{\mathrm{2}} } \\ $$$$\therefore{p}\left(\mathrm{2}\right)\:{satisfies}\:{p}\left({n}\right) \\ $$$${let}\:{us}\:{assume}\:{p}\left({n}\right)\:{is}\:{true}\:{for}\:{n}={k} \\ $$$$\Rightarrow{p}\left({k}\right):\frac{{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} +...+{a}_{{k}} }{{k}}\geqslant\sqrt[{{k}}]{{a}_{\mathrm{1}} {a}_{\mathrm{2}} {a}_{\mathrm{3}} ...{a}_{{k}} } \\ $$$${now}\:{we}\:{have}\:{to}\:{show} \\ $$$${p}\left({k}+\mathrm{1}\right):\frac{{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} +...+{a}_{{k}+\mathrm{1}} }{{k}+\mathrm{1}}\geqslant\sqrt[{{k}+\mathrm{1}}]{{a}_{\mathrm{1}} {a}_{\mathrm{2}} {a}_{\mathrm{3}} ...{a}_{{k}+\mathrm{1}} } \\ $$$${please}\:{help} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

i am hereby posting proof given in Higher Algebrs Bernard and child...

$${i}\:{am}\:{hereby}\:{posting}\:{proof}\:{given}\:{in}\:{Higher}\:{Algebrs} \\ $$$${Bernard}\:{and}\:{child}... \\ $$$$\: \\ $$$$ \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

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