Let's consider this expression

    a^4 - a^3b + b^4 - ab^3.



Transform it this way  

    a^4 - a^3b + b^4 - ab^3 =  +  =  =  = .



So, our starting expression is the product of two quadratic polynomials

      and  .



They both are positively defined; in other words, they never take negative values.


Therefore,   >= 0  for all values of "a" and "b".



It implies that the original expression is never negative 

    a^4 - a^3b + b^4 - ab^3 >= 0.



It means that

    a^4 + b^4 >= a^3b + ab^3,


which is what has to be proved.

Tutor ikleyn's answer is 100% correct.  I wrote down a slightly different proof and thought I'd share, nothing earth-shattering.



Please note that the problem wording should be "...a and b are nonnegative real numbers..."   as "positive x" implies x>0.



Assume, WLOG, 



Then a=b+d for some  





=    (1)



and

   (2)



Now subtract (2) from (1) to get: 





 

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