Prove $\prod\limits_{k=1}^N (1+0.8\cdot r_{k})-1\leq \left( \prod\limits_{k=1}^{N}(1+r_{k}) -1 \right)\cdot0.8$

Question

Today I was asked to make a compound interest calculation in two different ways. And from this real life application, arose an interesting inequality that I was able to verify empirically but not algebraically. The inequality is the basically following: $$\prod_{k=1}^N (1+0,8\cdot r_{k})-1\leq \left( \prod_{k=1}^N(1+r_{k}) -1 \right)\cdot0,8$$ where $r_k$ is a positive Real number. The factor 0,8 was used in the actual calculation, but I believe it could be relaxed to be any real number between 0 and 1.

I admit that I am clueless on how to even start proving this inequality. Any hint is very welcome.

Answer 1

Your conjecture about the factor $0.8$ is correct. We can prove the following:

Consider $N \geq 1$ positive real numbers $r_1, r_2, \ldots, r_N$ and a scalar $r\in [0,1]$. Then the following inequality holds:

$$\prod_{k=1}^N(1+r\cdot r_k)-1 \leq r\cdot \left[\prod_{k=1}^N(1+r_k)-1\right]$$

I'm going to prove this using induction.

Case $N=1$ is just an identity. The case $N=2$, is:

$$(1+r\cdot r_1)(1+r\cdot r_2)-1\leq r\cdot \left[(1+r_1)(1+r_2)-1\right]$$

Expanding and canceling out similar terms, this is equivalent with:

$$(1-r)\cdot r_1r_2\geq 0$$

which is obvious. Now suppose that for some $n > 2$, we have:

$$\prod_{k=1}^n(1+r\cdot r_k)-1 \leq r\cdot \left[\prod_{k=1}^n(1+r_k)-1\right]$$

and we need to show

$$\prod_{k=1}^{n+1}(1+r\cdot r_k)-1 \leq r\cdot \left[\prod_{k=1}^{n+1}(1+r_k)-1\right]$$

Let $P = \displaystyle\prod_{k=1}^n(1+r_k)$. The assumption gives us that:

$$\prod_{k=1}^n(1+r\cdot r_k) \leq 1+r\cdot \left[P-1\right]$$

so

$$\prod_{k=1}^{n+1}(1+r\cdot r_k) = (1+r\cdot r_{n+1})\cdot \prod_{k=1}^n(1+r\cdot r_k) \leq (1+r\cdot r_{n+1})\left[1+r\cdot (P-1)\right]$$

Therefore, to complete the induction step, it is enough to show:

$$(1+r\cdot r_{n+1})\left[1+r\cdot (P-1)\right]-1\leq r\cdot \left[(1+r_{n+1})P-1\right]$$

and after some factorizations, this is equivalent with

$$r\cdot r_{n+1}(1-r)(P-1)\geq 0$$

which is obviously true because $r\in [0,1]$ and $P \geq 1$.

Answer 2

Let $$f(x) = \prod_{k=1}^N (1 + x r_k) - 1 - \Big(\prod_{k=1}^N (1+r_k) - 1\Big)x.$$ Clearly, $f''(x) \ge 0$ on $[0, 1]$. Thus, $f(x)$ is convex on $[0, 1]$. Also, $f(0) = f(1) = 0$. Thus, for any $0 \le x \le 1$, we have $f(x) = f((1-x)\cdot 0 + x\cdot 1) \le (1-x)f(0) + xf(1) = 0$. We are done.