prove or disprove $\ln{(1+x^2)}\cdot\ln{(1+y^2)}\ge \ln^2{(1+xy)}$

Question

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

Answer 1

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,

$$\begin{align}\lim_{x\to\infty}\left(\ln {\left(1+x^2\right)}\times \ln {\left(1+\frac {1}{x^2}\right)}\right)=0<\ln ^2 2.\end{align}$$

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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

$$\begin{align}\lim_{x\to\infty}\left(\ln {\left(1+x^2\right)}\times \ln {\left(1+\frac {1}{x^2}\right)}\right) &=\lim_{u\to\infty}\left(u\times \ln {\left(1+\frac {1}{e^u-1}\right)}\right)\\ &=\lim_{u\to\infty }\frac{\ln {\left(1+\frac {1}{e^u-1}\right)}}{\frac 1u}\\ &=\lim_{u\to\infty}\frac{-\frac{1}{e^u-1}}{-\frac {1}{u^2}}\\ &=\lim_{u\to\infty}\frac{u^2}{e^u-1}\\ &=2\lim_{u\to\infty}\frac{u}{e^u}\\ &=0.\end{align}$$

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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,

$$\begin{align}\lim_{x\to\infty}\left(\ln {\left(1+x^2\right)}\times \ln {\left(1+\frac{1}{x^2}\right)}\right)=0<\ln ^2 2.\end{align}$$

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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.

Answer 2

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.

Answer 3

(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:

$$ \ln (1+x^2) \times \ln (1+y^2) \leq \left( \frac{ \ln (1+x^2) + \ln (1+y^2) } { 2 } \right)^2 \geq \left( \ln ( 1+xy) \right)^2$$