Comparison between my formula and the number 1

Question

It is known that the real numbers $a>0$, $b>0$, and $0<\theta<1$, then is the following inequality correct? $$ \frac{a^\theta\cdot b^{1-\theta}}{\theta\cdot a+(1-\theta)\cdot b}\leq 1 $$ And why? Thanks in advance.

Answer 1

Since $\log x$ is concave, we have $$ \log(\theta a+(1-\theta)b)\geq \theta\log a+(1-\theta)\log b $$ for all $a,b>0$ and $\theta\in(0,1)$. Exponentiating both sides shows that $$ \theta a+(1-\theta)b\geq a^{\theta}b^{1-\theta}$$ which is equivalent to the stated inequality.