Inequality for sums of reciprocals

Question

Prove or disprove :: If $a_0,\ldots,a_n \geq m \geq 1$ then $\sum_{i=0}^n \frac{1}{a_i} \leq n + m / \prod_{i=0}^n a_i$.

[[ This is a generalization I formulated from this question :: Inequality Of Four Variables. ]]

Thank-you,

Moses

Answer 1

I have a proof by induction on $n$. Note that verifying the base case $n=0$ is trivial from $m \geq 1$.

Let us assume that the inequality is true for $n-1$,

Thus $\displaystyle \sum_{i=0}^{n-1} \frac{1}{a_i} \leq n-1 + \dfrac{m}{\prod_{i=0}^{n-1} a_i}$

Let us add $\frac1{a_n}$ on both sides and denote $\lambda_k = \prod_{i=0}^{k} a_i$. Note that $\lambda_{n-1}a_n = \lambda_n$. We get

$\displaystyle \sum_{i=0}^{n-1} \frac{1}{a_i} + \dfrac1{a_n} \leq n-1 + \dfrac{m}{\lambda_{n-1}} + \dfrac1{a_n}$

So to prove our statement it suffices to show

$n-1 + \dfrac{m}{\lambda_{n-1}} + \dfrac1{a_n} \leq n + \dfrac{m}{\lambda_{n}}$

By taking common denominators and using $\lambda_{n-1}a_n = \lambda_n$,

$\lambda_{n-1} +a_nm - a_n\lambda_{n-1} < m$

Rearranging it, we have to prove:

$(\lambda_{n-1} - m)(1 - a_n) \leq 0$

It is given that $a_n \geq 1$ and therefore $(1 - a_n) \leq 0$.

It is also given that $m \geq 1$ and $a_i \geq m$, therefore we have $\lambda_{n-1} = \prod_{i=0}^{n-1} a_i \geq m^n \geq m$. This implies that $(\lambda_{n-1} - m) \geq 0$.

Since $(1 - a_n) \leq 0$ and $(\lambda_{n-1} - m) \geq 0$, we can multiply the two inequalities to obtain $(\lambda_{n-1} - m)(1 - a_n) \leq 0$.

Therefore by induction, we have proved our inequality for all natural numbers.

Answer 2

If $n=0$ I think the case is trivial, thus assume that $n>0$.

If $m\ge 2$ then $$\sum_{i=0}^n \frac{1}{a_i} \le \sum_{i=0}^n \frac{1}{m} = \frac{n+1}{m}\le n.$$

Thus we are reduced to proving that if $a_0,\ldots,a_n\ge 1$ then $$\sum_{i=0}^n \frac{1}{a_i}\le n+ \frac{1}{\prod_{i=0}^n a_i}.$$ Well, if $a_x,a_y\ge 2$ for some $x$ and $y$, then $$\sum_{i=0}^n \frac{1}{a_i} \le \sum_{i\not= x,y} \frac{1}{a_i}+\frac{1}{2}+\frac{1}{2} \le (n-1)+1 = n \le n+\frac{1}{\prod_{i=0}^n a_i}.$$ Thus we may assume that $a_x\ge 2$ for some $x$, but $a_i=1$ for all $i\not=x$. In this case $$\sum_{i=0}^n \frac{1}{a_i} = n+\frac{1}{a_x} = n+\frac{1}{\prod_{i=0}^n a_i}.$$

Answer 3

Take $$f(a_0,\ldots,a_n)=\left(\sum_{i=0}^n \frac{1}{a_i}\right)-\frac{m}{a_0a_1\cdots a_n}=\frac{1}{a_j}\underbrace{\left(1-\frac{m}{a_0a_1\cdots\hat{a_j}\cdots a_n}\right)}_{\text{nonnegative}}+c$$ where $c$ is constant with respect to $a_j$. $f$ is maximized when $a_j$ is minimal, i.e. $a_j=m$.

Hence the overall maximum of $f$ is at $f(m,m,\ldots, m)=\frac{n+1}{m}-\frac{m}{m^{n+1}}=\frac{n+1}{m}-m^{-n}$. If $m=1$, this is equal to $n$. We have $$\frac{\partial }{\partial m}\left(\frac{n+1}{m}-m^{-n}\right)=-\frac{n+1}{m^2}+\frac{n}{m^{n+1}}=\frac{n}{m^{n+1}}\underbrace{\left(1-\frac{n+1}{n}m^{n-1}\right)}_{\textrm{negative for }m\ge 1}$$

Hence $f(m,m,\ldots,m)$ achieves its global maximum at $m=1$.