Inequality involving $\log_2$ For $x \in [1,\infty)$, set $f(x) := \log_2(x+1)$. I'd like to prove that
$$\forall x,y \in [1,\infty), \space f(x)f(y) \geq f(xy). \tag{1}$$
I've tried to prove this by working with the function $g(x) := f(x)f(y) - f(xy)$ for fixed $y \in [1,\infty)$. My idea was to prove that $g' \geq 0$. However, this turned out to be equivalent to proving that
$$\frac{\log(y+1)}{(x+1)\log(2)} \geq \frac{y}{xy+1}, \tag{2}$$
where $\log$ means "natural logarithm". So far I'm not able to prove $(2)$. Any idea on all this would be appreciated.
 A: I'm here coming up with my idea which does not follow your idea.
Set $g(x)={\ln(1+xy)\over\ln(1+x)}={\log_2(1+xy)\over\log_2(1+x)}$, then it is sufficient to show that the maximum of $g(x)$ is $f(y)=\log_2(1+y)$. The idea is followed from the property of monotonely decrease of $g(x)$ when $x\geq1$. Then the maximum of $g(x)$ is $g(1)={\log_2(1+y)\over\log_22}=f(y)$. To show $g(x)$ is monotonely decreasing,
$$g'(x)={\frac{y}{1+xy}\ln(1+x)-\frac1{1+x}\ln(1+xy)\over\ln^2(1+x)}={y(1+x)\ln(1+x)-(1+xy)\ln(1+xy)\over(1+xy)(1+x)\ln^2(1+x)}$$
The denominator of $g'(x)$ is always positive and denote by $h(x)$ the numerator of $g'(x)$. We need to show $h(x)=y(1+x)\ln(1+x)-(1+xy)\ln(1+xy)\leq0$. Again
$$h'(x)=y\ln(1+x)-y\ln(1+xy)\leq0$$
indicates $h$ is monotonely decreasing, namely
$$h(x)\leq\max h(x)=h(1)=y\ln4-(1+y)\ln(1+y)\leq0$$
Thus, required inequality holds.

Recapitulate the whole procedure, we have
$$\begin{align}f(x)f(y)\geq f(xy)
&\Longleftarrow \frac{f(xy)}{f(x)}=g(x)\leq f(y)\\
&\Longleftarrow g(x)\leq g(1)=f(y)\\
&\Longleftarrow g'(x)=\frac{h(x)}{\ell_+(x)}\leq0\\
&\Longleftarrow h(x)\leq h(1)\leq0\\
&\Longleftarrow h'(x)\leq0\end{align}$$ 
A: Hint: let $x+1=2^a$ and $y+1=2^b$, thus $a,b\geq1$. Hence $x=2^a-1$ and $y=2^b-1$, so $xy+1=2^{ab}-2^a-2^b+2$.
 The inequality then reads $ab\geq\log(2^{a+b}-2^a-2^b+2)$.  Then remember that $ab=\log(2^{ab})$.  So we try to show
$$2+2^{a+b}\leq2^a+2^b+2^{ab}.$$
First I've shown $2^{a+b}\leq 2^b+2^{ab}$.  After some calculation the inequality is equivalent to $\dfrac{\ln(2^a-1)}{\ln(2^{a-1})}\leq b$. Now the lhs is decreasing and for $a=1$ equals (via Hospital) $2$.  So the inequality is shown for $a\geq1$ and $b\geq2$.  But I didn't succeed to show it in case $1\leq a,b\leq2$, especially when $a$ is nera $1$ and $b$ near $2$. 
