For $a\ge0$ and $u\ge0$, let 
$$q(u):=\ln Q(a+\sqrt u). 
$$
Then 
$$q_2(t):=q''(u)\frac{8 \sqrt{2 \pi } t^3 e^{\frac{1}{2} (a+t)^2} Q(a+t)^2}{a t+t^2+1}
=2 Q(a+t)-\frac{\sqrt{\frac{2}{\pi }} t e^{-\frac{1}{2} (a+t)^2}}{a t+t^2+1},
$$
where $t:=\sqrt u\ge0$. 
Next,
$$q_2'(t)=-\frac{\sqrt{\frac{2}{\pi }} e^{-\frac{1}{2} (a+t)^2} (a t+2)}{\left(a t+t^2+1\right)^2}<0
$$
and $q_2(\infty-)=0$. So, $q_2>0$ and hence $q''>0$. So, $q$ is convex. So, $(a+b\sqrt u)$ is [log convex][1] in $u\ge0$ for any $a,b\ge0$. 

Now the desired result follows for any positive $\varepsilon_i$'s in view of the [well-known fact][2] that the sum of log-convex functions is log convex. 


  [1]: https://en.wikipedia.org/wiki/Logarithmically_convex_function
  [2]: https://mathoverflow.net/questions/288445/mixtures-of-log-convex-functions-are-log-convex-a-reference