, 2006

It is easily seen that $f(x)$ is a continuous function on $[0, \infty)$. We shall prove the following assertions:

i) $f(x)$ is strictly increasing on $[0, \infty)$;

ii) $f(0) = \sqrt{ab}$, $\lim_{x \to \infty} f(x) = \frac{a+b}{2}$;

iii) for every $s$ with $0 < s < 1$, we have:
$$\sqrt{ab} < \left(\frac{a^s + b^s}{2}\right)^{\frac{1}{s}} \le \frac{a+b}{2}.$$
With these assertions, the intermediate value theorem of continuous functions proves the existence of a unique $\alpha$ such that $f(\alpha) = \left(\frac{a^s + b^s}{2}\right)^{\frac{1}{s}}$.

Proof of i):
$$f'(x) = -1 + \frac{2x+a+b}{2\sqrt{(a+x)(b+x)}} = \frac{(\sqrt{a+x}-\sqrt{b+x})^2}{2\sqrt{(a+x)(b+x)}} > 0.$$

Proof of ii):
It is evident that $f(0) = \sqrt{ab}$ and:
$$\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{-x^2 + (a+x)(b+x)}{x + \sqrt{(a+x)(b+x)}} = \lim_{x \to \infty} \frac{a+b+\frac{ab}{x}}{1+\sqrt{\left(\frac{a}{x}+1\right)\left(\frac{b}{x}+1\right)}} = \frac{a+b}{2}.$$

Proof of iii):
The first inequality is deduced from the A-G-mean inequality.
Put $m = \left(\frac{a^s + b^s}{2}\right)^{\frac{1}{s}}$, $x = \frac{a}{m}$, $y = \frac{b}{m}$. Clearly, $x^s + y^s = 2$ and Bernoulli inequality implies that:
$$x = (1 + x^s - 1)^{\frac{1}{s}} \ge 1 + \frac{x^s - 1}{s}$$
$$y = (1 + y^s - 1)^{\frac{1}{s}} \ge 1 + \frac{y^s - 1}{s}$$
(both equalities do not hold simultaneously). From these inequalities, by adding both left sides together and both right sides together, we get $x + y > 2$.