Stochastic order on weighted sum of iid random variables

$$X_i$$ are n iid random variables with CDF $$1_{[0,+\infty[}(x) \Phi(x)$$, i.e. it is a mixture between a half Gaussian and a delta in $$0$$, both with weight $$1/2$$.

I would like to show that, $$\forall a \in \mathbb{R}_+^n$$, $$\parallel a\parallel_ 2 = 1 \Rightarrow \forall t\in \mathbb{R}_+^*$$, $$p(\sum_{i=1}^n a_i X_i \leq t) \geq p(\sum_{i=1}^n \frac{1}{\sqrt{n}} X_i \leq t)$$

If there wasn't the delta and $$X_i$$ was following a half gaussian distribution, the result would be a direct application of the theorem 2 from Yu 2011. However it is valid for a random variable with distribution, which is not my case here, and I struggle to see from their demonstration whether the presence of the delta is causing an issue. Is the result still true in my case ? If it is, how can I show it ?

Yu, Y. (2011). Some stochastic inequalities for weighted sums. Bernoulli, 17(3). https://arxiv.org/abs/0910.0544

This conjecture does not hold, even for $$n=2$$.
Indeed, then the difference between the left-hand side of your inequality and its right-hand side is $$d(a):=g(a)-g(1/\sqrt2),$$ where $$\begin{equation*} g(a):=g_t(a):=P(aX_1+\sqrt{1-a^2}\,X_2\le t) \end{equation*}$$ for real $$t>0$$ (which may be considered fixed) and $$a\in(0,1)$$.
Let now $$t=2$$. Then $$d(1/\sqrt2)=0$$ and $$\begin{equation*} d'(a) = g'(a) =e^{-\frac{2}{a^2}} \Big(\frac{1}{2 \pi \sqrt{1-a^2}}-\frac{\sqrt{\frac{2}{\pi }}}{a^2}\Big)+\frac{e^{\frac{2}{a^2-1}} (a^2+2 \sqrt{2 \pi } a-1)}{2 \pi (1-a^2)^{3/2}}, \tag{10}\label{10} \end{equation*}$$ so that $$d'(1/\sqrt2)=0$$ and $$\begin{equation*} d''(1/\sqrt2)=\frac{16-40 \sqrt{\pi }}{e^4 \pi }=-0.320\ldots<0, \end{equation*}$$ which implies that $$d(a)<0$$ for $$t=2$$ and $$a$$ close enough to $$1/\sqrt2$$. $$\quad\Box$$
Details on \eqref{10}: Let $$Y,Y_,Y_2$$ be iid half Gaussian random variables, each with pdf $$f_Y$$ such that $$f_Y(y)=\frac{2}{\sqrt{2\pi}}\,e^{-y^2/2}\,1(y>0)$$ for real $$y$$ and cdf $$F_Y$$, so that $$F'_Y(y)=f_Y(y)$$ for real $$y>0$$. Then \begin{equation*} \begin{aligned} &g(a) \\ &=\frac14+\frac12\,P(aY_1\le t) +\frac12\,P(\sqrt{1-a^2}\,Y_2\le t) +\frac14\,P(aY_1+\sqrt{1-a^2}\,Y_2\le t) \\ &=\frac14+\frac12\,F_Y\Big(\frac ta\Big) +\frac12\,F_Y\Big(\frac t{\sqrt{1-a^2}}\Big) +\frac14\,\int_0^\infty F_Y\Big(\frac{t-\sqrt{1-a^2}\,y}a\Big)f_Y(y)\,dy. \end{aligned} \end{equation*} So, \begin{equation*} \begin{aligned} &g'(a) \\ &=\frac{-t}{2a^2}\,f_Y\Big(\frac ta\Big) +\frac{t a}{2(1-a^2)^{3/2}}\,f_Y\Big(\frac t{\sqrt{1-a^2}}\Big) \\ &+\frac1{4a^2}\,\int_0^\infty \Big(\frac y{\sqrt{1-a^2}}-t\Big)f_Y\Big(\frac{t-\sqrt{1-a^2}\,y}a\Big)f_Y(y)\,dy. \end{aligned} \end{equation*} The latter integral is elementary, yielding \eqref{10}.