We have that

\(\displaystyle\star{\ln{{\left({1}+{x}\right)}}}\leq{x}\)

\(\displaystyle\star{\ln{{\left({1}−{x}\right)}}}\leq−{x}\)

therefore

\(\displaystyle{\left({1}−{x}\right)}{\ln{{\left({1}−{x}\right)}}}+{\left({1}+{x}\right)}{\ln{{\left({1}+{x}\right)}}}\leq−{x}+{x}^{{2}}+{x}+{x}^{{2}}={2}{x}^{{2}}\)

\(\displaystyle\star{\ln{{\left({1}+{x}\right)}}}\leq{x}\)

\(\displaystyle\star{\ln{{\left({1}−{x}\right)}}}\leq−{x}\)

therefore

\(\displaystyle{\left({1}−{x}\right)}{\ln{{\left({1}−{x}\right)}}}+{\left({1}+{x}\right)}{\ln{{\left({1}+{x}\right)}}}\leq−{x}+{x}^{{2}}+{x}+{x}^{{2}}={2}{x}^{{2}}\)