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

Annette Sabin 2022-01-02 Answered
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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Expert Answer

deginasiba
Answered 2022-01-03 Author has 1904 answers
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
Answered 2022-01-04 Author has 2796 answers
HINT:
Note that for \(\displaystyle{t}{>}{0}\)
\(\displaystyle{\log{{\left({t}\right)}}}\leq{t}−{1}\)
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