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

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 :)

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What have you tried? –  Mark Bennet Sep 9 '12 at 21:01
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

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.

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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.

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