Question Number 179961 by yaojunon2t last updated on 05/Nov/22
$${f}\left({x}\right)\in\left[\mathrm{0},\mathrm{1}\right],\mathrm{1}\leqslant{f}\left({x}\right)\leqslant\mathrm{3} \\ $$$${how}\:{to}\:{prove}\:\mathrm{1}\leqslant\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{{f}\left({x}\right)}\leqslant\frac{\mathrm{4}}{\mathrm{3}}? \\ $$
Answered by Frix last updated on 06/Nov/22
$$\mathrm{just}\:\mathrm{a}\:\mathrm{few}\:\mathrm{thoughts} \\ $$$$\mathrm{if}\:{f}\left({x}\right)={c}\:\Rightarrow\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{cdx}={c}\:\mathrm{and}\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{dx}}{{c}}=\frac{\mathrm{1}}{{c}}\:\Rightarrow \\ $$$$\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{both}\:\mathrm{is}\:{c}×\frac{\mathrm{1}}{{c}}=\mathrm{1} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{allow}\:\mathrm{an}\:\mathrm{intervall}\:{p}\:\mathrm{with}\:{f}\left({x}\right)={qc} \\ $$$$\mathrm{we}\:\mathrm{get} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}−{p}} {\int}}{cdx}+\underset{\mathrm{1}−{p}} {\overset{\mathrm{1}} {\int}}{qcdx}=\left({pq}−{p}+\mathrm{1}\right){c} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}−{p}} {\int}}\frac{{dx}}{{c}}+\underset{\mathrm{1}−{p}} {\overset{\mathrm{1}} {\int}}\frac{{dx}}{{qc}}=\frac{\frac{{p}}{{q}}−{p}+\mathrm{1}}{{c}} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{product}\:\mathrm{is}\:\frac{\left(\mathrm{1}+{p}\left({q}−\mathrm{1}\right)\right)\left({q}−{p}\left({q}−\mathrm{1}\right)\right)}{{q}} \\ $$$$\mathrm{min}\:\frac{\left(\mathrm{1}+{p}\left({q}−\mathrm{1}\right)\right)\left({q}−{p}\left({q}−\mathrm{1}\right)\right)}{{q}}\:=\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{1}\leqslant\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{f}\left({x}\right){dx}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{dx}}{{f}\left({x}\right)}\:\blacksquare \\ $$$$ \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{try}\:\mathrm{the}\:\mathrm{following} \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}}{c}_{\mathrm{1}} {dx}+\underset{\frac{\mathrm{1}}{\mathrm{2}}} {\overset{\mathrm{1}} {\int}}{c}_{\mathrm{2}} {dx}=\frac{{c}_{\mathrm{1}} +{c}_{\mathrm{2}} }{\mathrm{2}} \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}}\frac{{dx}}{{c}_{\mathrm{1}} }+\underset{\frac{\mathrm{1}}{\mathrm{2}}} {\overset{\mathrm{1}} {\int}}\frac{{dx}}{{c}_{\mathrm{2}} }=\frac{{c}_{\mathrm{1}} +{c}_{\mathrm{2}} }{\mathrm{2}{c}_{\mathrm{1}} {c}_{\mathrm{2}} } \\ $$$$\mathrm{the}\:\mathrm{product}\:\mathrm{is}\:\frac{\left({c}_{\mathrm{1}} +{c}_{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}{c}_{\mathrm{1}} {c}_{\mathrm{2}} }={P} \\ $$$$\mathrm{now}\:\mathrm{let}'\mathrm{s}\:\mathrm{try}\:{P}>\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\frac{\left({c}_{\mathrm{1}} +{c}_{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}{c}_{\mathrm{1}} {c}_{\mathrm{2}} }=\frac{\mathrm{4}}{\mathrm{3}}+\lambda \\ $$$$\Rightarrow\:{c}_{\mathrm{2}} ={c}_{\mathrm{1}} \left(\mathrm{2}\lambda+\frac{\mathrm{5}}{\mathrm{3}}\pm\frac{\mathrm{2}\sqrt{\mathrm{9}\lambda^{\mathrm{2}} +\mathrm{15}\lambda+\mathrm{4}}}{\mathrm{3}}\right)={c}_{\mathrm{1}} {Q}\left(\lambda\right) \\ $$$$\mathrm{we}\:\mathrm{know}\:\mathrm{that}\:\mathrm{1}\leqslant{c}_{\mathrm{1},\:} {c}_{\mathrm{2}} \leqslant\mathrm{3}\:\Rightarrow\: \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\leqslant{Q}\left(\lambda\right)\leqslant\mathrm{3} \\ $$$$\mathrm{but}\:{Q}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{or}\:\mathrm{3}\:\mathrm{and}\:{Q}\left(\lambda\neq\mathrm{0}\right)<\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{or}\:>\mathrm{3} \\ $$$$\Rightarrow\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{f}\left({x}\right){dx}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{dx}}{{f}\left({x}\right)}\leqslant\frac{\mathrm{4}}{\mathrm{3}}\:\blacksquare \\ $$$$ \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{sure}\:\mathrm{if}\:\mathrm{this}\:\mathrm{is}\:\mathrm{enough}\:\mathrm{to}\:\mathrm{be}\:\mathrm{a}\:\mathrm{proof} \\ $$
Commented by yaojunon2t last updated on 07/Nov/22
$${thanks}\:{a}\:{lot} \\ $$