# A Little of Algebra for an Inequality, A Little of Calculus for a Generalization

### Problem

If Let $a,b,c$ be positive pairwise distinct real numbers, $n$ a positive integer. Prove that

$\displaystyle \frac{a^{n+1}-b^{n+1}}{a-b}\cdot\frac{b^{n+1}-c^{n+1}}{b-c}\cdot\frac{c^{n+1}-a^{n+1}}{c-a}\gt (n+1)^3(abc)^n.$

Note that, as $a,b,c$ converge to a common value, the inequality morphs into equality.

### Solution 1

The inequality is "separable" and can be solved a factor at a time. By the AM-GM inequality,

\displaystyle \begin{align} \frac{a^{n+1}-b^{n+1}}{a-b}&=\sum_{k=0}^na^kb^{n-k}\\ &\ge (n+1)\cdot\sqrt[n+1]{(ab)^{\frac{n(n+1)}{2}}}\\ \frac{b^{n+1}-c^{n+1}}{b-c}&\ge (n+1)\cdot\sqrt[n+1]{(bc)^{\frac{n(n+1)}{2}}}\\ \frac{c^{n+1}-a^{n+1}}{c-a}&\ge (n+1)\cdot\sqrt[n+1]{(ca)^{\frac{n(n+1)}{2}}}\\ \end{align}

The product of which is

\displaystyle \begin{align} &\frac{a^{n+1}-b^{n+1}}{a-b}\cdot\frac{b^{n+1}-c^{n+1}}{b-c}\cdot\frac{c^{n+1}-a^{n+1}}{c-a}\\ &\qquad\ge (n+1)^3\cdot\sqrt[n+1]{(ab)^{\frac{n(n+1)}{2}}(bc)^{\frac{n(n+1)}{2}}(ca)^{\frac{n(n+1)}{2}}}\\ &\qquad= (n+1)^3\cdot\sqrt[n+1]{(abc)^{n(n+1)}}\\ &\qquad= (n+1)^3(abc)^{n}.\\ \end{align}

Equality in

$\displaystyle \left(\sum_{k=0}^na^kb^{n-k}\right)\left(\sum_{k=0}^nb^kc^{n-k}\right)\left(\sum_{k=0}^nc^ka^{n-k}\right)\ge (n+1)^3(abc)^{n}$

whenever $a=b=c,$ with strict inequality, otherwise.

### Solution 2

Let $x=a/b$, $y=b/c$, and $z=c/a$. Noting that none of the three defined variables equals $1$, the AM-GM gives a strict inequality as

\displaystyle\begin{align} 1+x+x^2+\ldots+x^n &\gt (n+1)x^{n/2} \\ \frac{x^{n+1}-1}{x-1} &\gt (n+1)x^{n/2}.\end{align}

Multiplyling the other two cylic terms,

\displaystyle \begin{align} \frac{x^{n+1}-1}{x-1} \cdot \frac{y^{n+1}-1}{y-1} \cdot \frac{z^{n+1}-1}{z-1} &\gt (n+1)^3(xyz)^{n/2}=(n+1)^3 \\ \frac{(a/b)^{n+1}-1}{(a/b)-1}\cdot\frac{(b/c)^{n+1}-1}{(b/c)-1}\cdot\frac{(c/a)^{n+1}-1}{(c/a)-1} &\gt (n+1)^3 \\ \frac{a^{n+1}-b^{n+1}}{a-b}\cdot\frac{b^{n+1}-c^{n+1}}{b-c}\cdot\frac{c^{n+1}-a^{n+1}}{c-a} &\gt (n+1)^3(abc)^n. \end{align}

### Generalization

Assume $t\gt 1.$ Then for any positive real numbers $a,b,c,$ pairwise distinct,

$\displaystyle \frac{a^{t+1}-b^{t+1}}{a-b}\cdot\frac{b^{t+1}-c^{t+1}}{b-c}\cdot\frac{c^{t+1}-a^{t+1}}{c-a}\gt (t+1)^3(abc)^t.$

### Proof

First note that the inequality is equivalent to

$\displaystyle \left(\frac{1}{a-b}\int_b^ax^tdx\right)\left(\frac{1}{b-c}\int_c^bx^tdx\right)\left(\frac{1}{c-a}\int_a^cx^tdx\right)\gt (abc)^t.$

Similar to the original problem, the one at hand is also separable, and we can consider the three integrals one at a time. Note also that, since $x^t\gt 0$ for $x\gt 0,$ the expression $\displaystyle \frac{1}{a-b}\int_b^ax^tdx$ is always positive independent of whether $a\gt b$ or $b\gt a.$ Since the function $f(x)=x^t$ is convex, we can apply the Hermite-Hadamard inequality,

$\displaystyle f\left({\frac {a+b}{2}}\right)\leq {\frac {1}{b-a}}\int _{a}^{b}f(x)\,dx\leq {\frac {f(a)+f(b)}{2}}$

which yields (along with the AM-GM inequality)

\displaystyle \begin{align} \frac{1}{a-b}\int_b^ax^tdx &\ge \left(\frac{a+b}{2}\right)^t\ge\left(\sqrt{ab}\right)^t\\ \frac{1}{b-c}\int_c^bx^tdx &\ge \left(\frac{b+c}{2}\right)^t\ge\left(\sqrt{bc}\right)^t\\ \frac{1}{c-a}\int_a^cx^tdx &\ge \left(\frac{c+a}{2}\right)^t\ge\left(\sqrt{ca}\right)^t \end{align}

from which

$\displaystyle \left(\frac{1}{a-b}\int_b^ax^tdx\right)\left(\frac{1}{b-c}\int_c^bx^tdx\right)\left(\frac{1}{c-a}\int_a^cx^tdx\right)\ge (abc)^t.$

As before, equality is only achieved when $a,b,c$ converge to a common value.

### Acknowledgment

The problem has been originally posted by Dorin Marghidanu at the CutTheKnotMath facebook page and later commented on with a generalization (and its proof) by Leo Giugiuc.

Solution 2 is by Amit Itagi.