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](https://mathoverflow.net/questions/285515/a-problem-involving-the-error-function) that $y\_0$ exists and is unique on $[-a,\infty)$. Thanks!



Root problem involving error function
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](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&cad=rja&uact=8&ved=0ahUKEwj8j_ugtJXZAhUH42MKHe49APoQFggvMAE&url=http%3A%2F%2Femis.um.ac.ir%2Fjournals%2FJIPAM%2Fimages%2F157_05_JIPAM%2F157_05.pdf&usg=AOvVaw00cKlwNdtMkwYwenqbUP-n), $h$ 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](https://my.pcloud.com/publink/show?code=XZqDGz7ZcJjmeGOeFMmk6WEDbFMTMBukn500) and/or [its pdf image](https://my.pcloud.com/publink/show?code=XZEDGz7ZWO6wpcvwRLVeePfB0yg1vF6SfNv7).  

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](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&cad=rja&uact=8&ved=0ahUKEwj8j_ugtJXZAhUH42MKHe49APoQFggvMAE&url=http%3A%2F%2Femis.um.ac.ir%2Fjournals%2FJIPAM%2Fimages%2F157_05_JIPAM%2F157_05.pdf&usg=AOvVaw00cKlwNdtMkwYwenqbUP-n), $\frac1r=\frac GF$ 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](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&cad=rja&uact=8&ved=0ahUKEwj8j_ugtJXZAhUH42MKHe49APoQFggvMAE&url=http%3A%2F%2Femis.um.ac.ir%2Fjournals%2FJIPAM%2Fimages%2F157_05_JIPAM%2F157_05.pdf&usg=AOvVaw00cKlwNdtMkwYwenqbUP-n), $1/r(1-)$ is decreasing and hence $r(1-)$ is increasing, in $a>0$. Also, $r(1-)\to1$ as $a\to0+$. So, indeed $r(1-)>1$. 


