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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!

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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 $r(1-)>1$.

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  • $\begingroup$ You are exactly right. I had a typo. It should be that $y_0$ is on interval $[-a,\infty)$. In fact I had checked this inequality by plotting it. I'm sure the statement is true, but I don't know how to prove. $\endgroup$ – Jackie Lu Feb 8 '18 at 0:04
  • $\begingroup$ Where does DR come from? And I guess you meant DR<0? $\endgroup$ – Jackie Lu Feb 9 '18 at 2:45
  • $\begingroup$ @JackieLu : The introduction of $DR$ was perhaps the subtlest point in the proof. If you look at the expression for $R$ in the Mathematica notebook or its pdf image, you will see that the most difficult to analyze Erf[...] term in it has a nonconstant coefficient, say $c(a)$. If you divide $R$ by $c(a)$ and then differentiate, the Erf[...] term will disappear, but the subsequent analysis will be difficult, because of certain non-monotonicity. $\endgroup$ – Iosif Pinelis Feb 9 '18 at 15:09
  • $\begingroup$ Above comment continued: The trick that turns out to work well is to divide $R$, not by the entire coefficient $c(a)$, but by the product (say $c_1(a)$) of just two non-constant factors in $c(a)$. Then the derivative of $R/c_1(a)$ will still contain the Erf[...] term, with a certain coefficient, say $c_2(a)$. However, we divide the derivative of $R/c_1(a)$ by $c_2(a)$ -- denoting the result of this division by $DR$ -- and the derivative of $DR$ is, not only free of Erf[...], but also manifestly negative. This then implies $DR<0$ (thank you for spotting the typo there; it is now corrected). $\endgroup$ – Iosif Pinelis Feb 9 '18 at 15:12
  • $\begingroup$ I see. Thank you!!! $\endgroup$ – Jackie Lu Feb 10 '18 at 21:40

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