Log Concavity Inequality

Last updated Nov 1, 2022

Let $a, b > 0$ and let $0 \leq \lambda \leq 1$. Then the following inequality holds $$a^{\lambda}b^{1-\lambda} \leq \lambda a + (1-\lambda) b$$ with equality If and Only If $\lambda = 0$ or $1$, or $a=b$.

By Strict Concavity of Log, $$\begin{align*} &\lambda \log a + (1-\lambda) \log b \leq \log (\lambda a + (1-\lambda)b)\\ \Rightarrow& \log (a^{\lambda} b^{1-\lambda}) \leq \log(\lambda a + (1-\lambda) b)\\ \Rightarrow& a^{\lambda}b^{1-\lambda} \leq \lambda a + (1-\lambda) b \end{align*}$$ because the Exponential is Non-decreasing.

Equality certainly holds if $\lambda = 0,1$ since the inequality reduces to either $a \leq a$ or $b \leq b$. If $a=b$ then the inequality reduces to $a \leq a$. The other direction follows from Strict Concavity of Log and the definition of a Strictly Convex Function. It states that if $\lambda \in (0,1)$ and $a \neq b$, then the inequality is strict. Thus, by Contraposition, if equality holds, then either $\lambda \in {0,1}$ or $a=b$. $\blacksquare$