Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $a$ and $b$ be positive numbers, and $n \in \mathbb{N}$. Prove that (using Rearrangement Inequality)

$$(n+1)\left(a^{n+1}+b^{n+1}\right) \geq (a+b)\left(a^n+a^{n-1}b+\ldots+b^{n}\right). $$

Thanks :)

share|cite|improve this question
it is possible not have any idea. only calculations. – Iuli Sep 9 '12 at 21:05
As a fun fact, it is a statement of the average of $x^n$ over $[a,b]$. $$\frac{{{b^{n + 1}} + {a^{n + 1}}}}{{b + a}} \geqslant \frac{1}{{b - a}}\int\limits_a^b {{x^n}dx} $$ – Pedro Tamaroff Sep 9 '12 at 21:07
Multiply RHS with $a-b$ to obtain $(a+b)(a^{n+1}-b^{n+1})=a^{n+2}+a^{n+1}b-ab^{n+1}-b^{n+2}$ Do the same with the LHS to obtain $(n+1)(a^{n+2}-a^{n+1}b+b^{n+1}a-b^{n+2})$. Hence you actually want to compare $n\cdot(a^{n+2}-b^{n+2})$ against $(n+2)a b (a^n-b^n)$. Does that help? – Hagen von Eitzen Sep 9 '12 at 21:30
Iuli: Sorry but I do not understand your comment answering @MarkBennet's. Could you explain? – Did Sep 10 '12 at 7:19
up vote 3 down vote accepted

As the sequences $(a^{k+1},b^{k+1}),(a^{n-k},b^{n-k})$ are similarly sorted, therefore Rearrangement Inequality gives $a^{n+1} + b^{n+1} \ge a^{k+1}b^{n-k} + a^{n-k}b^{k}$.

The same inequality for $k-1$ is $a^{n+1} + b^{n+1} \ge a^{k}b^{n-k+1} + a^{n-k+1}b^{k-1}$, hence adding them gives $$a^{n+1} + b^{n+1} \ge (a+b)(a^{k}b^{n-k} + a^{n-k}b^{k})$$ Adding these inequalities for $k=0,1,2, \cdots, n $ gives the required.

share|cite|improve this answer

The inequality is obviously true if $a = b$, because we have $2(n+1)a^{n+1}\geq 2na^{n+1}$, which is true for $a\geq 0$. Now, if $a\neq b$, we have $a^n+a^{n-1}b+\dots+b^n=\displaystyle\frac{a^{n+1}-b^{n+1}}{a-b}$. Let's assume $a>b$ (otherwise we can simply inverse the roles of $a$ and $b$). The inequality to prove is equivalent to $(n+1)(a^{n+1}+b^{n+1})(a-b)\geq (a+b)(a^{n+1}-b^{n+1})$, or $(n+1)(a^{n+2}-b^{n+2} +ab(b^n - a^n))\geq a^{n+2}-b^{n+2}+ab(a^n-b^n)$, or again $n(a^{n+2}-b^{n+2})\geq nab(a^n-b^n)$, or $(a^{n+2}-b^{n+2})\geq ab(a^n-b^n)$, or $1-x^{n+2}\geq x(1-x^n)$ with $x = a/b$ (if $a=0$, then necessarily $b=0=a$ and this case has been covered previously).

Finally we need to prove that $\forall x\in [0,1), 1-x^{n+2}-x(1-x^n)=\varphi(x)\geq 0$. Now $\varphi'(x) = -(n+2)x^{n+1}-1+(n+1)x^n = -1 -((n+2)x - (n+1))x^n$

We can call $g(x) = ((n+2)x - (n+1))x^n$. We have $g'(x)=(n+2)x^n + n((n+2)x-(n+1))x^{n-1}=((n+1)(n+2)x - n(n+1))x^{n-1}$. $g'(x) = 0$ when $x = x_n = \frac{n}{n+2}$, so $g$ is maximized in $x_n$ and $g(x_n) = -(n/(n+2))^n< 1$, which proves that $\varphi'<0$, and thus $\varphi$ is decreasing, which finally achieves to prove the initial inequality.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.