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 ?
On
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
$$> \frac{1}{a+b+1/2}>\frac{1}{a+b+1}.$$
On
Set $a+b+1=1/c$, so $$ \frac{2a+2b+3}{2a+2b+1}= \frac{\frac{2}{c}+1}{\frac{2}{c}-1}=\frac{2+c}{2-c} $$ and the inequality is $$ \log\frac{2+c}{2-c}>c $$ for $0<c\le1$.
Consider the function $$ f(x)=\log(2+x)-\log(2-x)-x $$ defined over $[0,1]$. Then $f(0)=0$ and $$ f'(x)=\frac{1}{2+x}+\frac{1}{2-x}-1=\frac{x^2}{4-x^2}>0 $$ over the interval $(0,1)$. So the function is increasing.
$$\int_{a+b+1/2}^{a+b+3/2}\frac{dx}{x}=\log(\frac{2a+2b+3}{2a+2b+1})>\frac{1}{a+b+1}$$
This is true because $1/x$ is a strictly convex function. More generally, for any strictly convex $f$, we actually have:
$$\frac{1}{x_2-x_1}\int_{x_1}^{x_2}f(x)\,dx > f\left(\frac{x_1+x_2}{2}\right)$$