Menu Close

f-x-0-1-1-f-x-3-how-to-prove-1-0-1-f-x-dx-0-1-dx-f-x-4-3-

Tinku Tara 0 Comments
Pin




Question Number 179961 by yaojunon2t last updated on 05/Nov/22
f(x)∈[0,1],1≤f(x)≤3 how to prove 1≤∫_0 ^1 f(x)dx ∫_0 ^1 (dx/(f(x)))≤(4/3)?
$${f}\left({x}\right)\in\left[\mathrm{0},\mathrm{1}\right],\mathrm{1}\leqslant{f}\left({x}\right)\leqslant\mathrm{3} \\ $$$${how}\:{to}\:{prove}\:\mathrm{1}\leqslant\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{{f}\left({x}\right)}\leqslant\frac{\mathrm{4}}{\mathrm{3}}? \\ $$
Answered by Frix last updated on 06/Nov/22
just a few thoughts if f(x)=c ⇒ ∫_0 ^1 cdx=c and ∫_0 ^1 (dx/c)=(1/c) ⇒ the product of both is c×(1/c)=1 if we allow an intervall p with f(x)=qc we get ∫_0 ^(1−p) cdx+∫_(1−p) ^1 qcdx=(pq−p+1)c ∫_0 ^(1−p) (dx/c)+∫_(1−p) ^1 (dx/(qc))=(((p/q)−p+1)/c) and the product is (((1+p(q−1))(q−p(q−1)))/q) min (((1+p(q−1))(q−p(q−1)))/q) =1 ⇒ 1≤∫_0 ^1 f(x)dx∫_0 ^1 (dx/(f(x))) ■ now we try the following ∫_0 ^(1/2) c_1 dx+∫_(1/2) ^1 c_2 dx=((c_1 +c_2 )/2) ∫_0 ^(1/2) (dx/c_1 )+∫_(1/2) ^1 (dx/c_2 )=((c_1 +c_2 )/(2c_1 c_2 )) the product is (((c_1 +c_2 )^2 )/(4c_1 c_2 ))=P now let′s try P>(4/3) (((c_1 +c_2 )^2 )/(4c_1 c_2 ))=(4/3)+λ ⇒ c_2 =c_1 (2λ+(5/3)±((2(√(9λ^2 +15λ+4)))/3))=c_1 Q(λ) we know that 1≤c_(1, ) c_2 ≤3 ⇒ (1/3)≤Q(λ)≤3 but Q(0)=(1/3) or 3 and Q(λ≠0)<(1/3) or >3 ⇒ ∫_0 ^1 f(x)dx∫_0 ^1 (dx/(f(x)))≤(4/3) ■ I′m not sure if this is enough to be a proof
$$\mathrm{just}\:\mathrm{a}\:\mathrm{few}\:\mathrm{thoughts} \\ $$$$\mathrm{if}\:{f}\left({x}\right)={c}\:\Rightarrow\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{cdx}={c}\:\mathrm{and}\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{dx}}{{c}}=\frac{\mathrm{1}}{{c}}\:\Rightarrow \\ $$$$\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{both}\:\mathrm{is}\:{c}×\frac{\mathrm{1}}{{c}}=\mathrm{1} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{allow}\:\mathrm{an}\:\mathrm{intervall}\:{p}\:\mathrm{with}\:{f}\left({x}\right)={qc} \\ $$$$\mathrm{we}\:\mathrm{get} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}−{p}} {\int}}{cdx}+\underset{\mathrm{1}−{p}} {\overset{\mathrm{1}} {\int}}{qcdx}=\left({pq}−{p}+\mathrm{1}\right){c} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}−{p}} {\int}}\frac{{dx}}{{c}}+\underset{\mathrm{1}−{p}} {\overset{\mathrm{1}} {\int}}\frac{{dx}}{{qc}}=\frac{\frac{{p}}{{q}}−{p}+\mathrm{1}}{{c}} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{product}\:\mathrm{is}\:\frac{\left(\mathrm{1}+{p}\left({q}−\mathrm{1}\right)\right)\left({q}−{p}\left({q}−\mathrm{1}\right)\right)}{{q}} \\ $$$$\mathrm{min}\:\frac{\left(\mathrm{1}+{p}\left({q}−\mathrm{1}\right)\right)\left({q}−{p}\left({q}−\mathrm{1}\right)\right)}{{q}}\:=\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{1}\leqslant\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{f}\left({x}\right){dx}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{dx}}{{f}\left({x}\right)}\:\blacksquare \\ $$$$ \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{try}\:\mathrm{the}\:\mathrm{following} \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}}{c}_{\mathrm{1}} {dx}+\underset{\frac{\mathrm{1}}{\mathrm{2}}} {\overset{\mathrm{1}} {\int}}{c}_{\mathrm{2}} {dx}=\frac{{c}_{\mathrm{1}} +{c}_{\mathrm{2}} }{\mathrm{2}} \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}}\frac{{dx}}{{c}_{\mathrm{1}} }+\underset{\frac{\mathrm{1}}{\mathrm{2}}} {\overset{\mathrm{1}} {\int}}\frac{{dx}}{{c}_{\mathrm{2}} }=\frac{{c}_{\mathrm{1}} +{c}_{\mathrm{2}} }{\mathrm{2}{c}_{\mathrm{1}} {c}_{\mathrm{2}} } \\ $$$$\mathrm{the}\:\mathrm{product}\:\mathrm{is}\:\frac{\left({c}_{\mathrm{1}} +{c}_{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}{c}_{\mathrm{1}} {c}_{\mathrm{2}} }={P} \\ $$$$\mathrm{now}\:\mathrm{let}'\mathrm{s}\:\mathrm{try}\:{P}>\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\frac{\left({c}_{\mathrm{1}} +{c}_{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}{c}_{\mathrm{1}} {c}_{\mathrm{2}} }=\frac{\mathrm{4}}{\mathrm{3}}+\lambda \\ $$$$\Rightarrow\:{c}_{\mathrm{2}} ={c}_{\mathrm{1}} \left(\mathrm{2}\lambda+\frac{\mathrm{5}}{\mathrm{3}}\pm\frac{\mathrm{2}\sqrt{\mathrm{9}\lambda^{\mathrm{2}} +\mathrm{15}\lambda+\mathrm{4}}}{\mathrm{3}}\right)={c}_{\mathrm{1}} {Q}\left(\lambda\right) \\ $$$$\mathrm{we}\:\mathrm{know}\:\mathrm{that}\:\mathrm{1}\leqslant{c}_{\mathrm{1},\:} {c}_{\mathrm{2}} \leqslant\mathrm{3}\:\Rightarrow\: \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\leqslant{Q}\left(\lambda\right)\leqslant\mathrm{3} \\ $$$$\mathrm{but}\:{Q}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{or}\:\mathrm{3}\:\mathrm{and}\:{Q}\left(\lambda\neq\mathrm{0}\right)<\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{or}\:>\mathrm{3} \\ $$$$\Rightarrow\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{f}\left({x}\right){dx}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{dx}}{{f}\left({x}\right)}\leqslant\frac{\mathrm{4}}{\mathrm{3}}\:\blacksquare \\ $$$$ \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{sure}\:\mathrm{if}\:\mathrm{this}\:\mathrm{is}\:\mathrm{enough}\:\mathrm{to}\:\mathrm{be}\:\mathrm{a}\:\mathrm{proof} \\ $$
Commented by yaojunon2t last updated on 07/Nov/22
thanks a lot
$${thanks}\:{a}\:{lot} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *