There must be a mistake in your question since d/d=1
You corrected the mistake.
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#3
lendav_rott
232
10
multiplication comes before addition.
I see d/d-1 as \frac{d}{d} -1log 0 = who-knows-what
#4
maajdl
Gold Member
391
29
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 .