prove or disprove $\ln{(1+x^2)}\cdot\ln{(1+y^2)}\ge \ln^2{(1+xy)}$ let $x,y>0$.prove or disprove
$$\ln{(1+x^2)}\cdot\ln{(1+y^2)}\ge \ln^2{(1+xy)}$$
By this inequality it seem Cauchy-Schwarz inequality
$$(1+x^2)(1+y^2)\ge (1+xy)^2$$
But this is Log function.so How to prove it? Thanks
 A: It seems to me that the inequality is not correct.
Putting $y=\frac 1x$ we get,
$$\ln{\left(1+x^2\right)}\times\ln{\left(1+\frac {1}{x^2}\right)}\ge \ln^2{2}$$
But,
$$\begin{align}\lim_{x\to\infty}\left(\ln {\left(1+x^2\right)}\times \ln {\left(1+\frac {1}{x^2}\right)}\right)=0<\ln ^2 2.\end{align}$$

Evaluation of the limit:
Let, $1+x^2=e^u$. This implies, $1+\frac{1}{x^2}=1+\frac{1}{e^u-1}.$
Then, we have
$$\begin{align}\lim_{x\to\infty}\left(\ln {\left(1+x^2\right)}\times \ln {\left(1+\frac {1}{x^2}\right)}\right)
&=\lim_{u\to\infty}\left(u\times \ln {\left(1+\frac {1}{e^u-1}\right)}\right)\\
&=\lim_{u\to\infty }\frac{\ln {\left(1+\frac {1}{e^u-1}\right)}}{\frac 1u}\\
&=\lim_{u\to\infty}\frac{-\frac{1}{e^u-1}}{-\frac {1}{u^2}}\\
&=\lim_{u\to\infty}\frac{u^2}{e^u-1}\\
&=2\lim_{u\to\infty}\frac{u}{e^u}\\
&=0.\end{align}$$

But, we can calculate the limit faster.
Using the inequality,
$$\ln (1+u)<u, ~u>0$$
We have,
$$\begin{align}0&\le\ln\left(1+x^2\right)\ln\left(1+\frac{1}{x^2}\right)\\
&\le\ln\left(\left(1+x\right)^2\right)\ln\left(1+\frac{1}{x^2}\right)\\
&=2\ln\left(1+x\right)\ln\left(1+\frac{1}{x^2}\right)\\
&\le \frac{2x}{x^2}\\
&=\frac{2}{x}\end{align}$$
and $$\lim_{x\to \infty}\frac 2x=0.$$
Then, the Squeeze theorem tells us that,
$$\begin{align}\lim_{x\to\infty}\left(\ln {\left(1+x^2\right)}\times \ln {\left(1+\frac{1}{x^2}\right)}\right)=0<\ln ^2 2.\end{align}$$

This effective solution using the Squeeze theorem was made by @BarryCipra.
Since I found this method nicer and more useful, I added this to my answer.
A: The reverse is true:

Problem: Let $x, y > 0$. Prove that
$\ln(1 + x^2) \ln(1 + y^2) \le \ln^2(1 + xy)$.

Proof:
With the substitutions $x = \mathrm{e}^{u/2}$ and $y = \mathrm{e}^{v/2}$,
taking logarithm, it suffices to prove that
$$\ln \ln (1 + \mathrm{e}^{u}) + \ln \ln (1 + \mathrm{e}^{v})
\le 2 \ln \ln (1 + \mathrm{e}^{(u + v)/2})$$
which is true since
$g(w) = \ln \ln (1 + \mathrm{e}^{w})$ is concave:
$$g''(w) = -\frac{ \mathrm{e}^{w}( \mathrm{e}^{w} - \ln(1 + \mathrm{e}^{w}))}{(1 + \mathrm{e}^{w})^2\ln^2(1 + \mathrm{e}^{w} )} < 0$$
where $\mathrm{e}^{w} - \ln(1 + \mathrm{e}^{w}) > 0$ is easy to prove by taking derivative.
We are done.
A: (This is wrong. Leaving this up here for now in case someone tries this approach.)
Hint: Your CS inequality demonstrates that
$$\ln (1+x^2) + \ln (1+y^2) \geq 2 \ln (1+xy).$$
Now apply AM-GM to $ \ln (1+x^2) \times \ln (1+y^2)$.
However, that doesn't work. We get:
$$ \ln (1+x^2) \times \ln (1+y^2) \leq \left( \frac{ \ln (1+x^2) + \ln (1+y^2) } { 2 } \right)^2 \geq \left( \ln ( 1+xy) \right)^2$$
