I ran into this problem in my research:
Let $y_0$ be the root of
$$-(y+a)e^{y^2}\mathit{erfc}(y)+\frac{b}{\sqrt{\pi}}=0$$
on interval $[-a,\infty)$, while $a>0$ and $0<b<1$.
How can I show
$$y_0\leq \frac{a(b-2)+\sqrt{a^2b^2+2b(1-b)}}{2(1-b)}?$$
It is already known that $y_0$ exists and is unique on $[-a,\infty)$. Thanks!
Your conjecture is correct. Indeed, let \begin{equation*} h(y):=\sqrt\pi\,(y+a)e^{y^2}\text{erfc}(y)=\frac{f(y)}{g(y)},\quad f(y):=\sqrt\pi\,\text{erfc}(y),\quad g:=f/h. \end{equation*} Then the "derivative ratio" \begin{equation*} \rho(y):=\frac{f'(y)}{g'(y)}=\frac{2 (a + y)^2}{1 + 2 a y + 2 y^2} \end{equation*} is up-down in $y>0$, that is, for some $c\in[0,\infty]$, $\rho$ increases on $(0,c]$ and decreases on $[c,\infty)$. (Thus, being up-down includes the possibilities of being up (that is, everywhere increasing) and of being down (that is, everywhere decreasing).) Also, $f(\infty-)=0=g(\infty-)$. So, by Proposition 4.3 in L’HOSPITAL-TYPE RULES FOR MONOTONICITY, $h$ is up-down. Also, $h(-a)=0$ and $h(\infty-)=1$. So, for each $b\in(0,1)$ there is a unique root $y_0>-a$ of the equation \begin{equation*} h(y_0)=b, \end{equation*} and this is the same $y_0$ as in the OP's question. Moreover, it follows that $h$ is increasing on $(0,y_0]$.
We have to show that
$$y_0\overset{\text(?)}\le y_1:=\frac{a(b-2)+\sqrt{a^2b^2+2b(1-b)}}{2(1-b)}$$
for $a>0$, $0<b<1$.
If $y_1<y_0$, then $h(y_1)<h(y_0)$, since $h$ is increasing on $(0,y_0]$. So, it is enough to show that $h(y_1)\ge h(y_0)[=b]$, that is,
\begin{equation*}
r:=r(b):=r(a,b):=h(y_1)/b\overset{\text(?)}\ge1.
\end{equation*}
We have
\begin{equation*}
\frac1r=\frac GF,\quad F:=F(b):=\text{erfc}(y_1),\quad G:=G(b):=F/r.
\end{equation*}
Consider the "derivative ratio"
\begin{equation*}
r_1:=r_1(b):=\frac{G'}{F'}
\end{equation*}
and
\begin{multline*}
d_1:=d_1(b):=r'_1(b)
\frac{\sqrt{b ((a^2-2) b+2)}}{(1-b)^2} (1 - b + a^2 b - a \sqrt{b (2 + (-2 + a^2) b)})^3 \\
=
2 \sqrt{b \left(\left(a^2-2\right) b+2\right)} \left(a^4 b (2 b+1)+a^2 \left(-5 b^2+3
b+2\right)+2 (b-1)^2\right) \\
-a \left(6 \left(a^2+1\right) b+2 \left(2 a^4-7 a^2+6\right)
b^3+\left(2 a^4+8 a^2-21\right) b^2+3\right).
\end{multline*}
Details of all calculations can be seen in the
Mathematica notebook and/or its pdf image.
Using a computer algebra (CA) program, one verifies that $d_1$ equals $r'_1(b)$ in sign, and so, $r'_1(b)>0$ iff $d_1>0$ iff $0<a<\frac{\sqrt{5}}{2}\ \&\ \frac{-a^2-4}{4 \left(a^2-2\right)}-\frac{1}{4}
\sqrt{\frac{a^2-8}{a^2-2}}<b<1$.
So, the "derivative ratio" $r_1$ is down-up in $b\in(0,1)$. Also, $F>0$ and (by CA) $F'<0$.
So, by the last row of Table 1.2 in L’HOSPITAL-TYPE RULES FOR MONOTONICITY, $\frac1r=\frac GF$ is down-up-down and hence $r$ is up-down-up. But $r'(0+)=-\infty<0$, and so, $r$ is actually down-up.
The behavior of $r'(b)$ near $b=1$ is more complicated. Let \begin{equation*} R:=R(a):=r'(1-),\quad DR:=\frac{2 a^3 e^{-a^2-\frac{1}{4 a^2}}}{8 a^6+4 a^4-2 a^2+1}\, \frac d{da}\Big(R\, \frac{8 ea^5}{2 a^2 + 1}\Big), \end{equation*} \begin{equation*} \frac{d}{da}\,DR =-\frac{16 a^6 \left(32 a^{10}+144 a^8+208 a^6+96 a^4+50 a^2+35\right) e^{-a^2-\frac{1}{4 a^2}+1}}{\left(2 a^2+1\right)^3 \left(8 a^6+4 a^4-2 a^2+1\right)^2}<0. \end{equation*} We see that $DR$ decreases in $a>0$. Also, $DR\to0$ as $a\to0+$. So, $DR<0$, so that $R\, \frac{8 ea^5}{2 a^2 + 1}$ decreases. Also, $R\to0$ as $a\to0+$. So, $R<0$; that is, $r'(1-)<0$. So, $r$ is (not just down-up) but plainly down; that is, $r$ is decreasing in $b\in(0,1)$.
It remains to show that $r(1-)>1$. We have \begin{equation*} r(1-)=\frac{F_1}{G_1}, \quad F_1:=1-\text{erf}\left(\frac{1}{2 a}-a\right),\quad G_1:=F_1/r(1-). \end{equation*} The "derivative ratio" \begin{equation*} \frac{G_1'}{F_1'}=\frac{-4 a^4+2 a^2+1}{2 a^2+1} \end{equation*} is decreasing. Also, $F_1(0+)=0=G_1(0+)$. So, by Proposition 4.1 in L’HOSPITAL-TYPE RULES FOR MONOTONICITY, $1/r(1-)$ is decreasing and hence $r(1-)$ is increasing, in $a>0$. Also, $r(1-)\to1$ as $a\to0+$. So, indeed $r(1-)>1$.
