Question Number 45353 by pieroo last updated on 12/Oct/18
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{p}\left(\mathrm{n}\right)=\frac{\boldsymbol{\mathrm{a}}_{\mathrm{1}} +\boldsymbol{\mathrm{a}}_{\mathrm{2}} +…+\boldsymbol{\mathrm{a}}_{\mathrm{n}} }{\mathrm{n}}\:\geqslant\:^{\mathrm{n}} \sqrt{\boldsymbol{\mathrm{a}}_{\mathrm{1}} \boldsymbol{\mathrm{a}}_{\mathrm{2}} …\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} } \\ $$$$\forall\:\mathrm{n}\:\in\boldsymbol{\mathrm{N}} \\ $$
Commented by pieroo last updated on 12/Oct/18
$$\mathrm{please}\:\mathrm{help} \\ $$
Commented by Kunal12588 last updated on 12/Oct/18
$${trying}\:{PMI}\left({Induction}\right) \\ $$
Commented by pieroo last updated on 14/Oct/18
$$\mathrm{I}\:\mathrm{still}\:\mathrm{need}\:\mathrm{help}\:\mathrm{urgently}\:\mathrm{please} \\ $$
Answered by Kunal12588 last updated on 12/Oct/18
$${p}\left({n}\right):\frac{{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} +…+{a}_{{n}} }{{n}}\geqslant\sqrt[{{n}}]{{a}_{\mathrm{1}} {a}_{\mathrm{2}} {a}_{\mathrm{3}} …{a}_{{n}} } \\ $$$${p}\left(\mathrm{1}\right):\frac{{a}_{\mathrm{1}} }{\mathrm{1}}={a}_{\mathrm{1}} \:\:\:\:\:\:\:\:\:\:\:\sqrt[{\mathrm{1}}]{{a}_{\mathrm{1}} }={a}_{\mathrm{1}} \\ $$$$\therefore{p}\left(\mathrm{1}\right)\:{satisfies}\:{p}\left({n}\right) \\ $$$${p}\left(\mathrm{2}\right):\frac{{a}_{\mathrm{1}} +{a}_{\mathrm{2}} }{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\sqrt[{\mathrm{2}}]{{a}_{\mathrm{1}} {a}_{\mathrm{2}} }=\sqrt{{a}_{\mathrm{1}} }\sqrt{{a}_{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{{a}_{\mathrm{1}} }−\sqrt{{a}_{\mathrm{2}} }\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\Rightarrow\:\frac{{a}_{\mathrm{1}} +{a}_{\mathrm{2}} −\mathrm{2}\sqrt{{a}_{\mathrm{1}} }\sqrt{{a}_{\mathrm{2}} }}{\mathrm{2}}\geqslant\mathrm{0} \\ $$$$\Rightarrow\frac{{a}_{\mathrm{1}} +{a}_{\mathrm{2}} }{\mathrm{2}}\geqslant\sqrt{{a}_{\mathrm{1}} }\sqrt{{a}_{\mathrm{2}} } \\ $$$$\therefore{p}\left(\mathrm{2}\right)\:{satisfies}\:{p}\left({n}\right) \\ $$$${let}\:{us}\:{assume}\:{p}\left({n}\right)\:{is}\:{true}\:{for}\:{n}={k} \\ $$$$\Rightarrow{p}\left({k}\right):\frac{{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} +…+{a}_{{k}} }{{k}}\geqslant\sqrt[{{k}}]{{a}_{\mathrm{1}} {a}_{\mathrm{2}} {a}_{\mathrm{3}} …{a}_{{k}} } \\ $$$${now}\:{we}\:{have}\:{to}\:{show} \\ $$$${p}\left({k}+\mathrm{1}\right):\frac{{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} +…+{a}_{{k}+\mathrm{1}} }{{k}+\mathrm{1}}\geqslant\sqrt[{{k}+\mathrm{1}}]{{a}_{\mathrm{1}} {a}_{\mathrm{2}} {a}_{\mathrm{3}} …{a}_{{k}+\mathrm{1}} } \\ $$$${please}\:{help} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18
$${i}\:{am}\:{hereby}\:{posting}\:{proof}\:{given}\:{in}\:{Higher}\:{Algebrs} \\ $$$${Bernard}\:{and}\:{child}… \\ $$$$\: \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18
