Geometric mean vs arithmetic mean in general case

Question

The following holds? Well known problem?

For any $a \in (0, 1), p, q \in [0, 1]$,

$$ p^aq^{1 - a} \leq ap + (1 -a)q. $$

When $a = 1/2$, this holds and is a very famous problem, but how about $a \in (0, 1)$ in general?

Answer 1

We can rewrite this as $$a\ln p+(1-a)\ln q\le\ln\left(ap+(1-a)q\right)$$ or $$af(p)+(1-a)f(q)\le f\left(ap+(1-a)q\right)\tag1$$ for $f(x)=\ln x$. A function $g$ on an interval $I$ for which $$ag(p)+(1-a)g(q)\ge g\left(ap+(1-a)q\right)\tag2$$ $(2)$ holds for all $p$, $q\in I$ and $a\in[0,1]$ is called convex. A sufficient condition for convexity is that $g''(x)\ge0$. Here the inequality goes the other way, so (1) says that $-f$ is convex, or we say that $f$ is concave. As $-f''(x)=1/x^2$ this is true.

Jensen's inequality extends this to several summands: $$\sum_{i=1}^n a_i g(p_i)\ge g\left(\sum_{i=1}^n a_i p_i\right)$$ when $g$ is convex on an interval $I$, $p_i\in I$, $a_i\ge0$ and $\sum a_i=1$. One case is the weighted AM/GM inequality $$p_1^{a_1}p_2^{a_2}\cdots p_n^{a_n}\le a_ip_1+\cdots+a_np_n$$ for $p_i>0$ and $a_i$ satisfying the above conditions.

Answer 2

For an intuitive understanding of the relation between the weighted AM-GM inequality and the "plain" AM-GM one, consider the case of rational $a=\frac{m}{n} \in (0,1)$ so that $m,n \in \mathbb{N} \,\mid \,0 \lt m \lt n\,$, where the weighted AM-GM inequality can be derived from AM-GM in a very elementary way:

$$ \begin{align} ap + (1 -a)q = \frac{m \,p + (n-m)\,q}{n} \,&=\, \frac{\overbrace{p+p+\cdots+p}^{m\;\text{times}} \;+\; \overbrace{q+q+\cdots q}^{n-m \;\text{times}}}{n} \\[5px] &\ge\; \sqrt[n]{\smash[b]{\underbrace{p \cdot p \,\cdots\, p}_{m\;\text{times}} \,\cdot\, \underbrace{q\cdot q \cdots q}_{n-m \;\text{times}}}} = p^{\frac{m}{n}}\, q^{\frac{n-m}{n}}=p^a\,q^{1-a} \end{align} $$

Note that the inequality holds for any $p, q \ge 0\,$, not retricted to $p,q \in [0,1]$.