# Prove (1−x)\ln(1−x)+(1+x)\ln(1+x)\leq2x^2\text{ for }0<x<1

Prove $$\displaystyle{\left({1}−{x}\right)}{\ln{{\left({1}−{x}\right)}}}+{\left({1}+{x}\right)}{\ln{{\left({1}+{x}\right)}}}\leq{2}{x}^{{2}}\ \text{ for }\ {0}{ < }{x}{ < }{1}$$

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deginasiba
We have that
$$\displaystyle\star{\ln{{\left({1}+{x}\right)}}}\leq{x}$$
$$\displaystyle\star{\ln{{\left({1}−{x}\right)}}}\leq−{x}$$
therefore
$$\displaystyle{\left({1}−{x}\right)}{\ln{{\left({1}−{x}\right)}}}+{\left({1}+{x}\right)}{\ln{{\left({1}+{x}\right)}}}\leq−{x}+{x}^{{2}}+{x}+{x}^{{2}}={2}{x}^{{2}}$$
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Melinda McCombs
HINT:
Note that for $$\displaystyle{t}{>}{0}$$
$$\displaystyle{\log{{\left({t}\right)}}}\leq{t}−{1}$$