Is it true that $ S(x)^2 \geq S'(x)\int_{-\infty}^{x}S(t) \, dt$, where $S(x) = A\gamma(x) + \gamma(x-C)$ and $\gamma$ is the standard Gaussian? For $C\geq0$ and $A\geq1$ define the function
$$
S(x) = A\gamma(x)+\gamma(x-C)
$$
where $\gamma$ denotes the standard Gaussian distribution in $\mathbb R$, $\gamma(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}$. Some numerical simulations suggest that the following inequality holds true:
$$
S(x)^2 \geq S'(x)\int_{-\infty}^{x}S(t) \, dt \qquad \forall x \in [0,C/2].
$$
I find it hard to prove this. Do you have any idea how to show this inequality?
 A: You want to show
$$
S(x)^2\geq S'(x)\int^{x}_{-\infty}S(t)dt\tag 0
$$
Let$$f(x)=\int^{x}_{-\infty}S(t)dt.\tag 1$$
Then your inequality is equivalent with the statement that $y(x)=\log(f(x))$ is strictly convex $\downarrow$ in $[0,C/2]\subset (-\infty,x_1]$, $C/2<x_1=x_1(A,C)\approx C/2$. That is because $f(x)>0$ in $(-\infty,x_1]$ and$$\frac{d^2}{dx^2}\log(f(x))=\frac{f''(x)f(x)-(f'(x))^2}{f(x)^2}=\frac{S'(x)\int^{x}_{-\infty}S(t)dt-S(x)^2}{\left(\int^{x}_{-\infty}S(t)dt\right)^2}.$$
Also for small $x\in
(x_1,x_2)$, $x_2=x_2(A,C)$, we have $y''(x)\geq 0$.
The convexity of $\log(f(x))$ for $x>>1$ (actualy $x>x_2$) follows from the fact that $f(x)$ is positive (as mentioned above) strictly inceasing and have limit  $$\lim_{x\rightarrow+\infty}f(x)=\int^{+\infty}_{-\infty}S(t)dt=1+A<\infty.$$Hence for $x>>1$ we have $y(x)$ convex $\downarrow$.
However I use a different approach.
Lemma 1. Assume that $h(x)$ is continuous in $[a,b]$, $a<b$. Then holds$$h(x)\geq0,\forall x\in[a,b]\Leftrightarrow\int^{y}_{x}h(t)dt\geq0,\forall x\leq y\textrm{ such }x,y\in[a,b]$$Proof. It is well known that if $h(t)\geq0$, then $\int^{y}_{x}h(t)dt\geq0$, for all $x\leq y$ with $x,y\in[a,b]$ . For the opposite we have that $\int^{x}_{a}h(t)dt\geq 0$, for all $x\in [a,b]$. Then assume that exists $x_0\in[a,b]$ such that $h(x_0)<0$. Then from the continouity of $h(x)$ we can assume that exist $c,d:a\leq c<d\leq b$ such that $h(x)<0$ in $[c,d]$. Then $a\leq c<d\leq b$ and $\int^{d}_{c}h(t)dt<0$, which is imposible.
Lemma 2. Assume that $f(x)\in C^{2}(D)$, $D=[a,b]$, $a<b$ and $f(x),f'(x)>0$, $\forall x\in D$. Then
$$
f(w)\leq f(z)\exp\left(\frac{f'(x)}{f(x)}(w-z)\right)\textrm{, }\forall b\geq w\geq z\geq x\geq a\tag 2
$$
iff
$$
f(x)f''(x)\leq (f'(x))^2\textrm{, }\forall x\in[a,b].\tag 3
$$
Proof.
For all $t\in[a,b]$
$$
f(t)f''(t)\leq(f'(t))^2\Leftrightarrow \frac{f''(t)}{f'(t)}\leq\frac{f'(t)}{f(t)}\Leftrightarrow
$$
$$
\int^{y}_{x}\frac{f''(t)}{f'(t)}dt\leq\int^{y}_{x}\frac{f'(t)}{f(t)}dt\Leftrightarrow 
$$
$$
\log\left(f'(y)\right)-\log\left(f'(x)\right)\leq \log\left(f(y)\right)-\log\left(f(x)\right)\Leftrightarrow
$$
$$
\log\left(\frac{f'(y)}{f(y)}\right)\leq\log\left(\frac{f'(x)}{f(x)}\right)\Leftrightarrow\frac{f'(y)}{f(y)}\leq\frac{f'(x)}{f(x)},
$$
for all $x\leq y$ such that $x,y\in[a,b]$. Now assume $w\geq z\geq y$. Then for the function
$$h(y)=\frac{f'(y)}{f(y)}-\frac{f'(x)}{f(x)}$$It holds $h(y)\leq 0$, for all $y\geq x$ and using Lemma 1, we get
$$
\log(f(w))-\log(f(z))\leq \frac{f'(x)}{f(x)}(w-z).
$$
Hence when $w\geq  z$  and $z,w\in[x,b]$, for all $x\in[a,b]$ we get
$$
\frac{\log(f(w))-\log(f(z))}{w-z}\leq \frac{f'(x)}{f(x)}.
$$
But since $f'(y)>0$ we get finaly
$$
\left|\log(f(w))-\log(f(z))\right|\leq \left|\frac{f'(x)}{f(x)}\right|\left|w-z\right|,
$$
for all $w\geq z\geq x$, $a\leq x\leq b$. Hence
$$
f(w)\leq f(z)\exp\left(\frac{f'(x)}{f(x)}(w-z)\right),\tag 4
$$
for all $w,z,x\in[0,C/2]: C/2\geq w\geq z\geq x\geq 0$. QED
The function $f(x)$ of (1) is positive in $[0,C/2]$. If we prove that
$$
\frac{f(w)}{f(z)}-\frac{1}{2}\left(\frac{f'(x)}{f(x)}\right)^2(w-z)^2-\frac{f'(x)}{f(x)}(w-z)-1\leq 0\tag 5 
$$
we are done (note that $e^t\geq t^2/2+t+1$, $\forall t\geq 0$).

