let-a-b-c-d-gt-1-f-a-b-c-d-a-continuous-function-for-which-a-b-such-that-a-a-f-x-dx-b-b-f-x-dx-a-c-then-prove-a-b-x-f-x-dx-

Question Number 219956 by hardmath last updated on 04/May/25

let a,b,c,d > 1 f : [a , b] → [c , d] a continuous function for which ∃λ ∈ (a , b) such that a ∫_a ^( 𝛌) f(x) dx + b ∫_b ^( 𝛌) f(x) dx ≥ a + c then prove ∫_a ^( b) (x/(f(x))) dx ≤ ((1/a) + (1/b)) ((b^2 −a^2 −2)/2)

$$\mathrm{let}\:\:\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\:>\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{f}\::\:\left[\mathrm{a}\:,\:\mathrm{b}\right]\:\rightarrow\:\left[\mathrm{c}\:,\:\mathrm{d}\right] \\ $$$$\mathrm{a}\:\mathrm{continuous}\:\mathrm{function} \\ $$$$\mathrm{for}\:\mathrm{which}\:\:\exists\lambda\:\in\:\left(\mathrm{a}\:,\:\mathrm{b}\right) \\ $$$$\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{a}\:\int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\lambda}} \:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{dx}\:+\:\mathrm{b}\:\int_{\boldsymbol{\mathrm{b}}} ^{\:\boldsymbol{\lambda}} \:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{dx}\:\geqslant\:\mathrm{a}\:+\:\mathrm{c} \\ $$$$\mathrm{then}\:\mathrm{prove} \\ $$$$\int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \:\frac{\mathrm{x}}{\mathrm{f}\left(\mathrm{x}\right)}\:\mathrm{dx}\:\leqslant\:\left(\frac{\mathrm{1}}{\mathrm{a}}\:+\:\frac{\mathrm{1}}{\mathrm{b}}\right)\:\frac{\mathrm{b}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} −\mathrm{2}}{\mathrm{2}} \\ $$

Answered by MrGaster last updated on 04/May/25

let λ⊂(a,b)satisfy:∫_(n−1) ^( n) (dx/(f(x))) ≤ (2/(n + 1)) ∵f(x)≥c>1⇒∫_a ^b (x/(f(x)))dx≤(1/c)∫_a ^b x dx=(1/c)∙((b^2 −a^2 )/2) Combined with the topicd conitions the existence λ k maes: a∫_a ^λ cdx+b∫_λ ^b c dx=c(a(λ−a)+b(b−λ))≥a+c λ=b,The left formula is ac(b−a)∧c=((a+c)/(a(b−a)))When the equation isd satisfie the target integral is: ∫_a ^b (x/c)=((b^2 −a^2 )/(2c)) Substitute c=((a+c)/(a(b−a+)) and get: ∫_a ^b (x/(f(x)))≤((1/a)−(1/b))((b^2 −a^2 −2)/2) Therefore the originall formua can be proved

$$\mathrm{let}\:\lambda\subset\left({a},{b}\right)\mathrm{satisfy}:\int_{\boldsymbol{\mathrm{n}}−\mathrm{1}} ^{\:\boldsymbol{\mathrm{n}}} \:\frac{\mathrm{dx}}{\mathrm{f}\left(\mathrm{x}\right)}\:\leqslant\:\frac{\mathrm{2}}{\mathrm{n}\:+\:\mathrm{1}}\: \\ $$$$\because{f}\left({x}\right)\geq{c}>\mathrm{1}\Rightarrow\int_{{a}} ^{{b}} \frac{{x}}{{f}\left({x}\right)}{dx}\leq\frac{\mathrm{1}}{{c}}\int_{{a}} ^{{b}} {x}\:{dx}=\frac{\mathrm{1}}{{c}}\centerdot\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{Combined}\:\mathrm{with}\:\mathrm{the}\:\mathrm{topicd} \\ $$$$\mathrm{conitions}\:\mathrm{the}\:\mathrm{existence}\:\lambda\:\mathrm{k} \\ $$$$\mathrm{maes}: \\ $$$${a}\int_{{a}} ^{\lambda} {cdx}+{b}\int_{\lambda} ^{{b}} {c}\:{dx}={c}\left({a}\left(\lambda−{a}\right)+{b}\left({b}−\lambda\right)\right)\geq{a}+{c} \\ $$$$\lambda={b},\mathrm{The}\:\mathrm{left}\:\mathrm{formula}\:\mathrm{is}\:{ac}\left({b}−{a}\right)\wedge{c}=\frac{{a}+{c}}{{a}\left({b}−{a}\right)}\mathrm{When}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{isd} \\ $$$$\mathrm{satisfie}\:\mathrm{the}\:\mathrm{target}\:\mathrm{integral}\:\mathrm{is}: \\ $$$$\int_{{a}} ^{{b}} \frac{{x}}{{c}}=\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{c}} \\ $$$$\mathrm{Substitute}\:{c}=\frac{{a}+{c}}{{a}\left({b}−{a}+\right.}\:\mathrm{and}\:\mathrm{get}: \\ $$$$\int_{{a}} ^{{b}} \frac{{x}}{{f}\left({x}\right)}\leq\left(\frac{\mathrm{1}}{{a}}−\frac{\mathrm{1}}{{b}}\right)\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} −\mathrm{2}}{\mathrm{2}} \\ $$$$\mathrm{Therefore}\:\mathrm{the}\:\mathrm{originall} \\ $$$$\mathrm{formua}\:\mathrm{can}\:\mathrm{be}\:\mathrm{proved} \\ $$

Commented by hardmath last updated on 04/May/25

$$ \\ $$Thank you, my dear, valuable magical mathematician, it's nice to see your perfect solutions.

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