Exploring an Inequality

Introduction

I have attempted to give a mathematical proof of the following inequality,

$$ \frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+d}+\frac{d}{1+a}\le 2, $$

where $a,\ b,\ c$ and $d$ are all in the range $[0,\ 1]$, but no success. So in this note I explore the above inequality using the power of computation.

Exploring

Define function

$$ f(a,\ b,\ c,\ d) = \frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+d}+\frac{d}{1+a}, $$

where $a,\ b,\ c$ and $d$ are all in the range $[0,\ 1]$.

Method 1

Calculate values of the above function for some given values of $a$, $b$, $c$ and $d$, and then give a summary.

R code:

library(dplyr)
f <- function(a, b, c, d)
{a/(1+b) + b/(1+c) + c/(1+d) + d/(1+a)
}
n <- 10
df <- 
  expand.grid(a = seq(0, 1, length.out = n), 
              b = seq(0, 1, length.out = n),
              c = seq(0, 1, length.out = n), 
              d = seq(0, 1, length.out = n)) %>% 
  mutate(the_f_value = f(a, b, c, d))
(summary(df$the_f_value))

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   0.000   1.212   1.442   1.399   1.632   2.000

NB: To save running time, I set $n=10$; in practice $n$ should be large, say $n=100$.

Method 2

Use base::optim to find the maximum of the function $f(\cdot)$.

g <- function(x)
{x[1]/(1 + x[2]) + x[2]/(1 + x[3]) + x[3]/(1 + x[4]) + x[4]/(1 + x[1])
}
(re <- optim(par = c(0, 0, 0, 0), fn = g, method = "L-BFGS-B",
             lower = rep(0, 4), upper = rep(1, 4), 
             control = list(fnscale = -1)))

## $par
## [1] 1 1 1 1
## 
## $value
## [1] 2
## 
## $counts
## function gradient 
##        2        2 
## 
## $convergence
## [1] 0
## 
## $message
## [1] "CONVERGENCE: NORM OF PROJECTED GRADIENT <= PGTOL"

NB:

• To use base::optim, we must define $f$ in the way like g in the above.
• In the par, we set an initial value.
• control = list(fnscale = -1) is used because we want to find the maximum rather than the minimum.