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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 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 that the sum of log-convex functions is log convex.

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, $Q(a+b\sqrt u)$ is log convex 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 that the sum of log-convex functions is log convex.

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 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   that the sum of log-convex functions is log convex.

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 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 that the sum of log-convex functions is log convex.

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 in $u\ge0$ for any $a,b\ge0$.

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

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 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 that the sum of log-convex functions is log convex.