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Let be $0 and $f(x)=\ln(x)$. According to Lagrange's theorem there is at least one point such that: $f'(c) = \frac{f{(b)}-f{(a)}}{b-a}$

Prove that $\sqrt{ab} < c < \frac{a+b}{2}$

I got the derivative of $f(x)=\ln(x)$ and then replaced x by c. The same thing for $f{(a)}, f{(b)}$ replacing them by $\ln(a) , \ln(b)$, but still there is no obvious way of going from this point on. I need some support here. Thanks.

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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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    ver$y$ 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