Statement: Let f1,f2,⋯,fn are n functions who are positive in the region [a,b]. And p1,p2,⋯,pn be positive rational numbers such that p1+p2+⋯+pn=1. Thenk=1∏n[(∫abfk(x)dx)pk]≥∫ab[k=1∏n(fk(x))pk]dx
Proof:
k=1∑n⎣⎢⎢⎢⎡∫abfk(x)dxpkfk⎦⎥⎥⎥⎤≥k=1∏n⎣⎢⎢⎢⎡⎝⎜⎜⎜⎛∫abfk(x)dxfk(x)⎠⎟⎟⎟⎞pk⎦⎥⎥⎥⎤
⟹∫ab⎣⎢⎢⎢⎡k=1∑n⎣⎢⎢⎢⎡∫abfk(x)dxpkfk(x)⎦⎥⎥⎥⎤⎦⎥⎥⎥⎤dx≥∫ab⎣⎢⎢⎢⎡k=1∏n⎣⎢⎢⎢⎡⎝⎜⎜⎜⎛∫abfk(x)dxfk(x)⎠⎟⎟⎟⎞pk⎦⎥⎥⎥⎤⎦⎥⎥⎥⎤dx
⟹k=1∑n⎣⎢⎢⎢⎡∫abfk(x)dxpk∫abfk(x)dx⎦⎥⎥⎥⎤≥k=1∏n[(∫abfk(x)dx)pk]∫ab[k=1∏n(fk(x))pk]dx
⟹k=1∏n[(∫abfk(x)dx)pk]≥∫ab[k=1∏n(fk(x))pk]dx
#Calculus
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Comments
Nice!
Wonderful I really love the argument