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

2$\begingroup$ this is true iff it is true for a=1. $\endgroup$– Dima PasechnikFeb 20, 2017 at 18:50

3$\begingroup$ Even more interestingly, it is true iff it is true for a=1 and b=1. Gerhard "Likes Better Living Through Continuity" Paseman, 2017.02.20. $\endgroup$– Gerhard PasemanFeb 20, 2017 at 19:22
5 Answers
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 AMGM 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: $$ (a^6 + b^6)  (a^5 b + a b^5) = (ab)^2 (a^4 + ab^3 + a^2 b^2 + ab^3 + b^4) $$ 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) = (ab) (a^5  b^5), $$ and note that $ab$ has the same sign as $a^5b^5$, 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.]
Follows from Hölder's inequality (p=6, q = 6/5):
$ab^5 + ba^5 \le (a^6+b^6)^{1/6} (b^6+a^6)^{5/6}$
This is easily proven using the Rearrangement Inequality, which says that if we have two sequences of reals, and we are to pair them up in such a way as to maximize the sum of the products of the pairs, then we should pair them up according to size. From this, it immediately follows that $a^6 + b^6 \geq a^5b+ab^5$ (for consider the sequences $(a^5,b^5)$ and $(a,b)$...).

1$\begingroup$ Careful: This argument only works when $a$ and $b$ are nonnegative. $\endgroup$ Oct 30, 2018 at 3:23

$\begingroup$ Case with $a,b$ of opposite signs is trivial, while case with $a,b$ both negative reduces to positive values by simultaneous change of signs. $\endgroup$ Oct 30, 2018 at 10:03
This is a very very special case of Muirhead's inequality. It involves two multiexponents $\newcommand{\bZ}{\mathbb{Z}}$ $\alpha,\beta\in\bZ_{\geq 0}^n$ and the natural action of the symmetric group $S_n$ on $\newcommand{\bR}{\mathbb{R}}$ $\bR^n$. A permutation $\phi$ acts on the vector $x=(x_1,\dotsc, x_n)$ by permuting the coordinates
$$\phi\cdot x=(x_{\phi(1)},\dotsc, x_{\phi(n)}). $$
(Warning: this is a right action, even though we write the $\phi$ on the left.)
We say that $\alpha \in\bZ_{\geq 0}^n$ precedes $\beta \in\bZ_{\geq 0}^n$, and we write this $\alpha\prec \beta$, if $\alpha$ lies in the convex hull of the set
$$ S_n\cdot \beta :=\big\{\; \phi\cdot\beta;\;\;\phi\in S_n\;\big\}. $$
Given $\alpha\in\bZ^n_{\geq 0}$ and $x\in\bR^n_{\geq 0}$ we define
$$ M_\alpha(x)=\sum_{\phi\in S_n} (\phi\cdot x)^\alpha= \sum_{\phi\in S_n} \prod_{k=1}^n x^{\alpha_k}_{\phi(k)}. $$
Muirhead's inequality states that
$$ \alpha\prec \beta \;\Longleftrightarrow M_\alpha(x)\leq M_\beta(x),\;\;\forall x\in\bR^n_{\geq 0}. $$
The inequality in your question corresponds to the case
$$ n=2, \;\;\alpha=(5,1)\prec \beta=(6,0),\;\; x=(a,b).$$
For more details on the rich story behind Muirhead's inequality I refer to J. Michael Steele's wonderful book
An Introduction to the Art of Mathematical Inequalities. The CauchySchwarz Master Class, Cambridge University Press, 2004.

1$\begingroup$ Careful: This argument only works when $a$ and $b$ are nonnegative. $\endgroup$ Oct 30, 2018 at 3:23

1$\begingroup$ @darijgrinberg You are correct. The general case follows from the positive case $$M_{5,1}(a.b)\leq M_{5,1}(a,b)\leq M_{6,0}(a.b)=M_{6,0}(a,b).$$ $\endgroup$ Oct 30, 2018 at 9:13