Given $ye^{-{xe^{y-2x}}} = 2xe^{-x},$ where $x>\frac{1}{\sqrt{2}}, y<\sqrt{2}$. Show that $y=y(x)$ decreases.

Question

Suppose $y$ is defined by the following implicit equation: $ye^{-{xe^{y-2x}}} = 2xe^{-x},$ where $x,y\geq 0.$

I want to show that $y$ decreases as $x$ increases, when $x>\frac{1}{\sqrt{2}}$ and $y<\sqrt{2}$. Here is my work:

Suppose by contradiction that there exist some $x_1, x_2> \frac{1}{\sqrt{2}}$ such that $x_1<x_2 \implies y_1<y_2.$

From the above relation, we have
$$2x_1 = y_1e^{x_1(1-e^{y_1-2x_1})} \tag{1}$$ and
$$2x_2 = y_2e^{x_2(1-e^{y_2-2x_2})} .\tag{2}$$

Subtracting this results in the following:
$$2(x_2-x_1) = y_2e^{x_2(1-e^{y_2-2x_2})}- y_1e^{x_1(1-e^{y_1-2x_1})}.$$

From here, I am not sure how to proceed. The goal is to arrive at the contradiction $y_1>y_2$ with the assumption but getting this inequality seems somewhat challenging.

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Following @jean's comment:

Take the natural logarithm on both sides of $(1)$ and $(2)$, and subtract:

\begin{align*}\ln{x_2}-\ln{x_1} &= \ln{y_2}-\ln{y_1} + x_2(1-e^{y_2-2x_2})- x_1(1-e^{y_1-2x_1})\\ & = \ln{y_2}-\ln{y_1} + (x_2-x_1) + (e^{y_1-2x_1}-e^{y_2-2x_2}). \end{align*}

This looks much better compared to the exponential expression.

Answer 1

Some thoughts.

Fact 1. $1 - xy\mathrm{e}^{y - 2x} > 0$ on $(\frac{1}{\sqrt{2}}, \infty)\times (0, \sqrt{2})$.

Fact 2. For each fixed $x > \frac{1}{\sqrt{2}}$, the equation $ \ln y - x \mathrm{e}^{y-2x} = \ln 2 + \ln x - x$ has exactly one real solution $y \in (0, \sqrt{2})$.

$\phantom{2}$

Now, let $$F(x, y) := \ln y - x \mathrm{e}^{y-2x} - (\ln 2 + \ln x - x).$$ Clearly, $F(x, y)$ is continuously differentiable on $(\frac{1}{\sqrt{2}}, \infty)\times (0, \sqrt{2})$. By Fact 1, we have $$\frac{\partial F}{\partial y} = \frac{1}{y}\left(1 - xy\mathrm{e}^{y - 2x}\right) > 0.$$ By the Implicit Function Theorem, using Fact 2, the equation $F(x, y) = 0$ implicitly determines $y$ as a differentiable function of $x$, given that $x\in (\frac{1}{\sqrt{2}}, \infty)$.

Taking derivative with respect to $x$ on $\ln y - x \mathrm{e}^{y-2x} - (\ln 2 + \ln x - x) = 0$, we have $$\frac{1}{y} \cdot y'(x) - \mathrm{e}^{y-2x} - x \mathrm{e}^{y-2x}(y'(x) - 2) - \frac{1}{x} + 1 = 0$$ or $$\frac{1}{y}\left(1 - xy\mathrm{e}^{y - 2x}\right)y'(x) = \frac{1}{x} + (1 - 2x)\mathrm{e}^{y - 2x} - 1. \tag{1}$$

From (1), using Fact 1, it suffices to prove that $$\frac{1}{x} + (1 - 2x)\mathrm{e}^{y - 2x} - 1 < 0,$$ or $$\mathrm{e}^{y - 2x} > \frac{1/x - 1}{2x - 1}\tag{2}$$ given that $(x, y) \in (\frac{1}{\sqrt{2}}, \infty)\times (0, \sqrt{2})$ with $\ln y - x \mathrm{e}^{y-2x} = (\ln 2 + \ln x - x)$.

Clearly, we only need to prove the case that $\frac{1}{\sqrt{2}} < x < 1$. (2) is written as $$y > 2x + \ln \frac{1/x - 1}{2x - 1}. \tag{3}$$ Clearly, we only need to prove the case that $2x + \ln \frac{1/x - 1}{2x - 1} > 0$. Note that $2x + \ln \frac{1/x - 1}{2x - 1} < \sqrt{2}$ for all $\frac{1}{\sqrt{2}} < x < 1$ (easy). By Fact 2, we have $y < \sqrt{2}$.

Let $f(y) := \ln y - x \mathrm{e}^{y-2x} - (\ln 2 + \ln x - x)$. By Fact 1, we have $f'(y) > 0$ on $(0, \sqrt{2})$, given that $x \in (\frac{1}{\sqrt{2}}, \infty)$. Thus, we have $$f(y) > f\left(2x + \ln \frac{1/x - 1}{2x - 1}\right) \iff y > 2x + \ln \frac{1/x - 1}{2x - 1}$$ given that $2x + \ln \frac{1/x - 1}{2x - 1}\in (0, \sqrt{2})$, and $y \in (0, \sqrt{2})$. Using $f(y) = 0$, it suffices to prove that $$0 > f\left(2x + \ln \frac{1/x - 1}{2x - 1}\right),$$ or $$\ln \left(2x + \ln \frac{1/x - 1}{2x - 1}\right) - x \cdot \frac{1/x - 1}{2x - 1} - (\ln 2 + \ln x - x) < 0 \tag{4}$$ for all $x \in (\frac{1}{\sqrt{2}}, 1)$ with $2x + \ln \frac{1/x - 1}{2x - 1} > 0$.

Note that $- x \cdot \frac{1/x - 1}{2x - 1} - (\ln 2 + \ln x - x) < 1 - \ln 2$ for all $\frac{1}{\sqrt{2}} < x < 1$. We only need to prove the case that $\ln \left(2x + \ln \frac{1/x - 1}{2x - 1}\right) + (1 - \ln 2) \ge 0$, i.e. $2x + \ln \frac{1/x - 1}{2x - 1} \ge \frac{2}{\mathrm{e}}$ which implies $x < 4/5$. Thus, it suffices to prove (4) for $x \in (\frac{1}{\sqrt{2}}, \frac45)$.

Using $\ln (1 + u) \ge \frac{u}{2}\cdot \frac{2 + u}{1 + u}$ for all $-1 < u \le 0$, using $\frac{1/x - 1}{2x - 1} \in (0, 1)$, we have $$\ln \frac{1/x - 1}{2x - 1} \ge - \frac{(2x^2 - 2x + 1)(2x^2 - 1)}{2x(1 - x)(2x - 1)}.$$ It suffices to prove that $$g(x) := \ln \left(2x - \frac{(2x^2 - 2x + 1)(2x^2 - 1)}{2x(1 - x)(2x - 1)}\right) - x \cdot \frac{1/x - 1}{2x - 1} - (\ln 2 + \ln x - x) < 0.$$ (Note: $2x - \frac{(2x^2 - 2x + 1)(2x^2 - 1)}{2x(1 - x)(2x - 1)} > 0$ for all $x \in (\frac{1}{\sqrt{2}}, \frac45)$.)

We have $$g'(x) = \frac{(2x^2 - 1)(24x^6 - 80x^5 + 116x^4 - 102x^3 + 56x^2 - 17x + 2)}{x(1 - x)(2x - 1)^2(-12x^4 + 16x^3 - 4x^2 - 2x + 1)} < 0.$$ Also, we have $g(\frac{1}{\sqrt{2}}) = 0$. Thus, $g(x) < 0$ for all $x \in (\frac{1}{\sqrt{2}}, \frac45)$.

Answer 2

An answer, which needs at the end a complementary study to be fully rigorous.

First of all, it is clear that setting $y=2x$ in the implicit equation gives $0$. Therefore, it is not surprizing that the following graphical representation of the implicit equation produced by GeoGebra has 2 branches : the straight line with equation $y=2x$ in blue and a hyperbola-looking branch in red.

Fig. 1.

By assuming that $y \ne 2x$, we can transform the initial equation into the following implicit equation which is the equation of the red branch the one of interest :

$$\frac{\ln(y)-\ln(2x)}{e^{y}-e^{2x}}=x e^{-2x}\tag{1}$$

The question is now : how can we prove that $y$ as defined by (1) can be given a cartesian equation $y=f(x)$ and that this function is a decreasing function of $x$ ?

Remark : this functionality issue (the existence of a unique $y$ for a given $x$) is assumed to be evident in the question but is not in fact ; it will be adressed later on.

Let us transform the LHS of (1) by setting

$$X=e^{2x}, \ \ Y=e^y \ \ \iff \ \ 2x=\ln X, \ \ y=\ln Y$$

giving :

$$\frac{\ln(\ln(Y))-\ln(\ln(X))}{Y - X}=\frac{\ln X}{2X} \tag{2}$$

Important : the domain is now $X>1$ and $Y>1$.

Setting

$$\varphi(X) := (\ln \circ \ln)(X) \ \ \text{(iterate log.) and } \ \ \psi(X)= \frac{\ln X}{2X}, $$

(with $\varphi'(X) = \frac{1}{X \ln(X)}$) relationship (2) can be written :

$$\frac{\varphi(Y)-\varphi(X)}{Y - X}= \psi(X)\tag{3}$$

New solution :

Let us consider the LHS of (3) as a "rate of change" (geometricaly the slope of line segment joining known point $A(X,\varphi(X))$ to point $B(Y,\varphi(Y))$ to be constructed ; see fig. 2 below).

The derivative $\varphi'$ is a bijective function (indeed, it is strictly decreasing) from $(1,\infty)$ to $(0,\infty)$ ; therefore we have, for each $X$, a (unique) value $X_0$ such that

$$\varphi'(X_0)=\psi(X)\tag{4}$$

giving to (3) the form :

$$\frac{\varphi(Y)-\varphi(X)}{Y - X}= \varphi'(X_0)\tag{5}$$

We can apply the mean value theorem (or more exactly a version of this theorem) allowing to conclude, due to its concavity, that, for any $X$, there exists a unique $Y$ such that the LHS of (5) is equal to its RHS, establishing in particular the functionality $Y=F(X)$ of the relationship of $X$ and $Y$, with, either :

$$X < X_0 <Y \ or \ Y < X_0 < X\tag{6}$$

Now, let us use a geometrical argument to see the pertinence of our approach (still not a rigorous proof along the standard "canons" of analysis). The closer $X$ is to value $1$ (i.e. the closer $x$ is to $0$), the farthest we have to take point $B(Y,\varphi(Y))$, i.e., as $X \to 1$, $Y \to \infty$. In the other direction, when $X \to \infty$, $Y \to 1$. This establishes the asymptotes (see figure below where $f$ is for $\varphi$ and $g$ is for $\psi$).

Fig. 2 : Caution : the domain of variables is here $X>1, \ Y>1$. The red curve is the image of the red curve of Fig. 1. Please note that function $\psi$ has on $(1,\infty)$ a minimum value $0$ and a maximal value $\tfrac{1}{2e}$.

It remains to express rigorously the fact that $Y=F(X)$, not only has the right asymptotes, but is also decreasing ; this can almost surely be done in a rigorous way by bracketing the values (i.e., using inequalities "squeezing" the expressions like expression (6)). If I have some time, I will attempt to do it.