Cauchy's Mean Theorem/Proof 2

Proof
Let:


 * $\map f x = \ln x$

for $x > 0$.

With a view to apply Jensen's Inequality: Real Analysis: Corollary, we can show that $f$ is concave on $\openint 0 \infty$.

By Second Derivative of Concave Real Function is Non-Positive, it is sufficient to show that $\map {f''} x \le 0$ for all $x > 0$.

We have, by Derivative of Natural Logarithm:


 * $\map {f'} x = \dfrac 1 x$

We then have, by Derivative of Power:


 * $\map {f''} x = -\dfrac 1 {x^2}$

As $x^2 > 0$ for all $x > 0$, we have:


 * $\dfrac 1 {x^2} > 0$

Therefore:


 * $-\dfrac 1 {x^2} = \map {f''} x < 0$

so $f$ is indeed concave on $\openint 0 \infty$.

As $x_1, x_2, \ldots, x_n$ are all positive, they all lie in the interval $\openint 0 \infty$.

We therefore have, by Jensen's Inequality: Real Analysis: Corollary:


 * $\displaystyle \map \ln {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } \ge \frac {\sum_{k \mathop = 1}^n \lambda_k \map \ln {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}$

for real $\lambda_1, \lambda_2, \ldots, \lambda_n \ge 0$, with at least one of which being non-zero.

As $n$ is a positive integer, we have:


 * $\dfrac 1 n > 0$

We can therefore set:


 * $\lambda_i = \dfrac 1 n$

for $1 \le i \le n$.

This gives:

and:

We therefore have:


 * $\map \ln {A_n} \ge \map \ln {G_n}$

Note that for $x > 0$:


 * $\dfrac 1 x = \map {f'} x > 0$

Therefore, by Derivative of Monotone Function, $f$ is increasing on $\openint 0 \infty$.

We therefore have:


 * $A_n \ge G_n$