<s>[INST] Why is $\log\left(\frac{2a+2b+3}{2a+2b+1}\right)>\frac{1}{a+b+1}$? 
Why is $\log\left(\frac{2a+2b+3}{2a+2b+1}\right)>\frac{1}{a+b+1}$ ? with $a,b\ge 0$

If this is true (I don't know whether it is true, just inserted some values in wolfram alpha) then I can show that $\frac{1}{a+b+1}<\int_{a}^{a+1}\int_{b}^{b+1}\frac{1}{x+y}dxdy$ 
Is it possible to see it from Taylor ?
 [/INST] We have $$\frac{2a+2a+3}{2a+2b+1} = 1+ \frac{2}{2a+2b+1} = 1+\frac{1}{a+b+1/2}.$$
The Maclauren series for $\log(1-x)=-x-x^2/2-x^3/3-\cdots.$  So your left side 
$$=\log\left(1-\frac{-1}{a+b+1/2}\right)  = \frac{1}{a+b+1/2} - \frac{1}{2}\left(\frac{1}{a+b+1/2}\right)^2+\cdots . $$
The series is alternating so the above is
$$> \frac{1}{a+b+1/2}>\frac{1}{a+b+1}.$$
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