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 .