Question 1177684
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<pre>

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 = {{{a^3*(a-b)}}} + {{{b^3*(b-a)}}} = {{{(a-b)*(a^3-b^3)}}} = {{{(a-b)*(a-b)*(a^2 + ab + b^2)}}} = {{{(a-b)^2*(a^2+ab+b^2)}}}.



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

    {{{(a-b)^2}}}  and  {{{a^2 + ab + b^2}}}.



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


Therefore,  {{{(a-b)^2*(a^2+ab+b^2)}}} >= 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.
</pre>

At this point, the proof is completed.