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 ?

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.
