Is there $b$ depending only on $a$ such that $\prod_{i=1}^n \Gamma(\lambda_i a) \le b$ for all $\lambda_i >0$ s.t. $\sum \lambda_i = 1$?

Question

Consider the Gamma function $$ \Gamma(x)=\int_{0}^{\infty} t^{x-1} e^{-t} \mathrm d t \quad \forall x>0. $$

From this Wikipedia page, I get for all $x_{1}, x_{2}>0$ and $t \in[0,1]$, $$ \Gamma\left(t x_{1}+(1-t) x_{2}\right) \leq \Gamma\left(x_{1}\right)^{t} \Gamma\left(x_{2}\right)^{1-t} . $$

Now we fix $a>0$. Is there some upper bound $b$ depending only on $a$ such that $$ \prod_{i=1}^n \Gamma(\lambda_i a) \le b \quad \forall \lambda_1, \ldots, \lambda_n >0 \text{ s.t. }\lambda_1 + \ldots \lambda_n = 1? $$

Answer 1

No. Since $\Gamma(x)\sim \frac1x$ as $x\to 0$, taking $n=2$, $\lambda_1=\epsilon$ and $\lambda_2=1-\epsilon$ gives $$\prod_{i=1}^2\Gamma(\lambda_ia)=\Gamma(a\epsilon)\Gamma(a-a\epsilon)\sim \frac{\Gamma(a)}{a\epsilon},$$ which is arbitrarily large as $\epsilon\to 0$.