Does anybody have a proof of the concavity of the $\log{x}$ that does not use calculus?
$\endgroup$ 71 Answer
$\begingroup$A proof by Cauchy induction, which does not involve calculus, that the arithmetic mean is no less than the geometric mean is given here. Let $\alpha$ be any real number between $0$ and $1$. Then there is a sequence of natural numbers $m(n)$, with $0\leqslant m(n)\leqslant n$ ($n=1,2,...$ ), such that the rational numbers $m(n)/n$ converge to $\alpha$ as $n\to\infty$. For given $n$ and positive reals $x$ and $y$, consider the arithmetic and geometric means of the $n$ positive reals $x_1,...,x_n$, where $x_1=\cdots=x_{m(n)}=x$ and $x_{m(n)+1}=\cdots=x_n=y$. The AM-GM inequality for $x_1,...,x_n$ is $$\frac{x_1+\cdots+x_n}n\geqslant(x_1\cdots x_n)^{1/n},$$or $$\frac{m(n)}{n}x+\left(1-\frac{m(n)}n\right)y\geqslant x^{m(n)/n}y^{1-m(n)/n}.$$By the continuity of the arithmetical functions involved, we have in the limit as $n\to\infty$ that$$\alpha x +(1-\alpha)y\geqslant x^{\alpha}y^{1-\alpha}.$$Now take the logarithm, and we are home.
$\endgroup$ 4
