Inequality with symmetric polynomials [closed]

Question

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

Answer 1

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.]

Answer 2

Follows from Hölder's inequality (p=6, q = 6/5):

$ab^5 + ba^5 \le (a^6+b^6)^{1/6} (b^6+a^6)^{5/6}$

Answer 3

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 $(a,b)$...).

Answer 4

Also follows from Muirhead's inequality since $(6,0)$ majorizes $(5,1)$.

Answer 5

This is a very very special case of Muirhead's inequality. It involves two multi-exponents $\newcommand{\bZ}{\mathbb{Z}}$ $\alpha,\beta\in\bZ_{\geq 0}^n$ 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 Cauchy-Schwarz Master Class, Cambridge University Press, 2004.