Question Number 155264 by SANOGO last updated on 28/Sep/21
$${en}\:{utilisant}\:{l}'{integrale}\:{de}\:{cauchy}\:{schwarz} \\ $$$${l}'{integrale}\:\int_{{o}} ^{\mathrm{1}} \frac{{f}\left({x}\right)}{{x}+\mathrm{1}}{dx}\:\:\:\:{est}\:{majoree}\:{par}? \\ $$$$ \\ $$
Answered by puissant last updated on 28/Sep/21
$${D}'{apres}\:{Augustin}\:{luis}\:{Cauchy},\:{on}\:{a}: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{f}\left({x}\right)}{{x}+\mathrm{1}}{dx}\:\leqslant\:\left(\int_{\mathrm{0}} ^{\mathrm{1}} {f}^{\mathrm{2}} \left({x}\right){dx}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\Rightarrow\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{f}\left({x}\right)}{{x}+\mathrm{1}}{dx}\:\leqslant\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\int_{\mathrm{0}} ^{\mathrm{1}} {f}^{\mathrm{2}} \left({x}\right){dx}}\:… \\ $$
Commented by SANOGO last updated on 28/Sep/21
$${toujours}\:{le}\:{puissant}\: \\ $$