This looks like a better fit for Math Stackexchange, because it's the kind of thing one learns from Olympiad problem books . . . One standard approach that has not been mentioned yet: We may assume $a,b$ are both positive (if one is zero it's easy; if they're of opposite sign then ${\rm LHS} > 0 > {\rm RHS}$; and if both negative, change to $-a,-b$). Then by the AM-GM inequality we have $$ \frac56 a^6 + \frac16 b^6 \geq \bigl((a^6)^5 b^6\bigr)^{1/6} = a^5 b, $$ and likewise $\frac16 a^6 + \frac56 b^6 \geq ab^5$, whence the desired $a^6 + b^6 \geq a^5 b + a b^5$ follows.
Yet another possibility is to factor the difference: $$ (a^6 + b^6) - (a^5 b + a b^5) = (a-b)^2 (a^4 + ab^3 + a^2 b^2 + ab^3 + b^4) $$ and check that the last factor is nonnegative (e.g. it's $\left|(a-\rho b) \, (a-\rho^2 b)\right|^2$ where $\rho$ is a 5th root of unity).
Either way we see that equality holds iff $a=b$.