An inequality

'coz p > 1

therefore 1/p <1
But 1/p +1/q =1
so 1/p =1-1/q
1-1/q <1
i.e. 1/q<0 or q>0

Now, a≥0 and b≥0
so a^p≥0 and b^q≥0
and a^p/p<1 and b^q/q<0
adding these
a^p/p + b^q/q<1 (when a≠0,b≠0) >0 (1)
and ab>0 (when a≠0,b≠0) (2)
From (1) and (2)
a^p/p + b^q/q>ab (when a>0,b>0)
and a^p/p + b^q/q≥ab (when a≥0,b≥0)

But p>1 means 1p < 1

edited...:)

Jensen's or generalized AM-GM inequality

I could'nt understand why you can say a^p1p is < 0 ?

when we consider a^p>0 (i.e. a≠0)
we know that 1/p<1
from these we can say a^p/p <1

same question already solved here....

http://targetiit.com/iit-jee-forum/posts/inequalities-sikhna-chahte-ho-to-sikho-varna-koi-b-11546.html
Q10 #12

in that post i cldn't get hw it cld be said that q>1

take p=\frac{m}{n} and q=\frac{r}{s}
since p>1>0, we also have q>1>0
we can write \frac{a^p}{p}+ \frac{b^p}{q}=\frac{nra^p + msb^p}{mr}
now applying AM-GM to nr numbers = a^p and ms numbers =b^p
\frac{nra^p+msb^q}{nr+ms}\ge \sqrt[{nr+ms}]{a^{mr}.b^{mr}} ------------------------(1)
but since \frac{1}{p}+\frac{1}{q}=1
we have
nr+ms=mr
hence substituting in (1) we have
\frac{a^p}{p}+\frac{b^p}{p}\ge ab

oh sorry i didnt know it had been posted before