To prove: Log[1 + b] >= 2 b/(2 + b) for b>=0

Note that it's equal when b=0.

You take the derivative of

Log[1 + b] - 2 b/(2 + b), and get

b

^{2}/((1 + b) (2 + b)

^{2}). This is non-negative

for b>=0. QED.

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