Lagrange - Cauchy inequality
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real-analysis
1 Answers
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The direct approach works: translating everything in terms of $u=\frac{b-a}a\geqslant0$, the inequalities to prove become, after some simplifications, $$ u\leqslant(1+u)\log(1+u)\leqslant u(1+\tfrac12u). $$ Since these are equal at $u=0$, it suffices to consider the derivatives, which yields the sufficient condition $0\leqslant\log(1+u)\leqslant u$. This is obviously true for every $u\geqslant0$ hence you are done.
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0very nice and short proof. Thanks. In fact, i see that the trick is to reduce all the thing to a single variable. – 2012-05-28