How to prove the inequality $a^6+b^6 \geqslant ab^5+a^5b$ for all $a, b \in \mathbb R$?



Inequality with symmetric polynomials
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** $a=b$.


[*Added later*: simpler yet $-$ factor the difference as
$$
(a^6 + b^6) - (a^5 b + a b^5) = (a-b) (a^5 - b^5),
$$
and note that $a-b$ has the same sign as $a^5-b^5$, so their product
is nonnnegative, and zero *iff* $a=b$. Like most of the other proofs,
this generalizes to prove $a^{m+1} + b^{m+1} \geq a^m b + a b^m$
for all odd $m>0$, and for all real $m>0$ if $a,b \geq 0$,
with the same equality condition $a=b$ in either case.]


