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:
