How to prove $a\ln(1+x_1)+(1-a)\ln(1+x_2)\le \ln(1+x)$ for $a\in(0,1)$ and $x_1+x_2=x$

Question

I am trying to prove an inequality that:

$a\ln(1+x_1)+(1-a)\ln(1+x_2)\le \ln(1+x)$

for $a\in(0,1)$ and $x_1+x_2=x$. I am trying using the series expansion. But it looks complex. Is there any way I can use convexity or some property to prove this.

Answer 1

Assuming $x_1, x_2>0$. Since ln is concave, by Jensen's inequality, \begin{align*} a\ln (1+x_1) + (1-a)\ln(1+x_2) &\leq \ln (a(1+x_1) + (1-a)(1+x_2))\\ &= \ln(1+ax_1 + (1-a)x_2). \end{align*} Since ln is increasing, and $1+ax_1 + (1-a)x_2 \leq 1+x_1+x_2$ (this is where we need $x_1, x_2>0$), then \begin{equation*} \ln(1+ax_1+(1-a)x_2) \leq \ln(1+x_1+x_2) = \ln(1+x). \end{equation*}

Answer 2

Note that the inequality is not true for all $x_1+x_2=x$ in the domain.

For instance, with $x_1=x_2=-\frac13$ and $a = \frac 12$,

$$LHS = \frac12 \ln \frac23 + \frac12 \ln \frac23 > \ln \frac13 = RHS$$