Prove that $a^{3} b + a b^{3} < a^{4} + b^{4}$ .

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Solution

Assume $a^{3} b + b^{3} a < a^{4} + b^{4}$
$\Rightarrow a^{3} b + b^{3} a - a^{4} - b^{4} < 0$
$\Rightarrow a^{3} (b - a) + b^{3} (a - b) < 0$
$\Rightarrow (a - b) (b^{3} - a^{3}) < 0$
$\Rightarrow (a - b) (b - a) (a^{2} + b^{2} + a b) < 0$
$\Rightarrow (a - b)^{2} (a^{2} + b^{2} + a b) > 0$
We already know that $(a - b)^{2} > 0, (a^{2} + b^{2} + a b) > 0$
Hence our assumption $a^{3} b + b^{3} a < a^{4} + b^{4}$ is correct

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