Question Number 145073 by gsk2684 last updated on 02/Jul/21
$$\mathrm{let}\:\mathrm{a}_{\mathrm{1}} ,\mathrm{a}_{\mathrm{2}} ,…,\mathrm{a}_{\mathrm{n}} \:\mathrm{be}\:\mathrm{positive} \\ $$$$\mathrm{real}\:\mathrm{numbers}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\mathrm{a}_{\mathrm{1}} +\mathrm{a}_{\mathrm{2}} +…+\mathrm{a}_{\mathrm{n}} =\mathrm{1}\:\mathrm{then}\:\mathrm{find}\: \\ $$$$\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\mathrm{a}_{\mathrm{1}} ^{\mathrm{a}_{\mathrm{1}} } .\mathrm{a}_{\mathrm{2}} ^{\mathrm{a}_{\mathrm{2}} } ….\mathrm{a}_{\mathrm{n}} ^{\mathrm{a}_{\mathrm{n}} } \:? \\ $$
Commented by shuvam last updated on 02/Jul/21
$${are}\:{you}\:{teacher}? \\ $$
Commented by gsk2684 last updated on 02/Jul/21
$$\mathrm{yes} \\ $$
Answered by MJS_new last updated on 02/Jul/21
$$\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{r}^{{r}} \:=\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{let}\:{a}_{\mathrm{1}} \rightarrow\mathrm{1}\wedge{a}_{\mathrm{2}} ={a}_{\mathrm{3}} =…={a}_{{n}} \rightarrow\mathrm{0}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\:{n}}\leqslant{a}_{\mathrm{1}} ^{{a}_{\mathrm{1}} } {a}_{\mathrm{2}} ^{{a}_{\mathrm{2}} } …{a}_{{n}} ^{{a}_{{n}} } <\mathrm{1} \\ $$
Commented by gsk2684 last updated on 02/Jul/21
$$\mathrm{thank}\:\mathrm{you}\: \\ $$$$\mathrm{hou}\:\mathrm{could}\:\mathrm{you}\:\mathrm{write}\:\mathrm{minimum}\:\mathrm{as}\:\frac{\mathrm{1}}{\mathrm{n}} \\ $$$$\mathrm{kindly}\:\mathrm{explain}\:\mathrm{sir} \\ $$
Commented by MJS_new last updated on 02/Jul/21
$$\mathrm{minimum}\:\mathrm{at}\:{a}_{\mathrm{1}} ={a}_{\mathrm{2}} =…={a}_{{n}} =\frac{\mathrm{1}}{{n}} \\ $$$$\Rightarrow \\ $$$${a}_{{k}} ^{{a}_{{k}} } =\left(\frac{\mathrm{1}}{{n}}\right)^{\frac{\mathrm{1}}{{n}}} =\frac{\mathrm{1}}{\:\sqrt[{{n}}]{{n}}} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{\mathrm{1}}{\:\sqrt[{{n}}]{{n}}}\:=\left(\frac{\mathrm{1}}{\:\sqrt[{{n}}]{{n}}}\right)^{{n}} =\frac{\mathrm{1}}{{n}} \\ $$