Why correct this inequality log(d/d-1)>(1/d)?

[parisa]

Why correct this inequality,

log(d/(d-1))>1/d for d≥2?

 

[maajdl]

There must be a mistake in your question since d/d=1
You corrected the mistake.

 

[lendav_rott]

multiplication comes before addition.
I see d/d-1 as $$\frac{d}{d} -1$$log 0 = who-knows-what

 

[maajdl]

Well, check first that this inequality is correct for d=2.
Then prove that the function log(d/(d-1)) will continue to be > than 1/d for d>2.
You could do that by analysis the derivative.

Get insight by making a plot of these functions.

Alternatively, study the function log(d/(d-1))-1/d .
Calculate its value for d=2, and analyze its behavior for d>2 .

 

[CellCoree]

This inequality needs to be corrected because it does not accurately represent the relationship between the two variables. The inequality log(d/(d-1))>1/d implies that the logarithm of the ratio of d and (d-1) is greater than the reciprocal of d. However, this is not always true for all values of d. For d≥2, the inequality holds true, but for values of d less than 2, the inequality is actually reversed. This is due to the fact that the logarithm function is not linear and has different properties for different values of d. Therefore, it is important to correct this inequality to ensure that it accurately reflects the relationship between the two variables for all values of d.