Prove $a^\alpha b^{1-\alpha} \le \alpha a + (1 - \alpha)b, \; a,b > 0,\; 0 < \alpha < 1$ I have no idea how to do this. Any help would be appreciated. The chapter I'm on is about differentiation and the mean value theorem.
Prove $a^\alpha b^{1-\alpha} \le \alpha a + (1 - \alpha)b, \; a,b > 0,\; 0 < \alpha < 1$

This result (in exactly the same notation as here) is discussed at length in Appendix to Hardy's "A Course of Pure Mathematics". Hardy mentions a very good approach based on Mean Value Theorem. If $a = b$ then both sides are equal to $a$. So let $a < b$. Then we will show that the strict inequality $$a^{\alpha}b^{1 - \alpha} < \alpha a + (1 - \alpha) b$$ holds. Clearly this inequality can be written as $$a^{\alpha}b^{1 - \alpha} - a < (1 - \alpha)b + a(\alpha - 1) = (1 - \alpha)(b - a)$$ i.e. $$b^{1 - \alpha} - a^{1 - \alpha} < (1 - \alpha)(b - a)a^{-\alpha}$$ This is easily established via mean value theorem. Clearly if $f(x) = x^{1 - \alpha}$ then $f'(x) = (1 - \alpha)x^{-\alpha}$ so that $f(b) - f(a) = (b - a)f'(c)$ for some $a < c < b$. This means that $b^{1 - \alpha} - a^{1 - \alpha} = (b - a)(1 - \alpha)c^{-\alpha}$. Since $-\alpha < 0$ it follows that $c^{-\alpha} < a^{-\alpha}$ and hence $$b^{1 - \alpha} - a^{1 - \alpha} = (b - a)(1 - \alpha)c^{-\alpha} < (b - a)(1 - \alpha)a^{-\alpha}$$ and we are done.
