Inequality involving $f(x)=e^x-x^2/2$ with $f'(a)=f'(b)$ ($a\ne b$)

Question

I'm currently working on a high-school level mathematical problem and have developed a solution approach, but I'm stuck at the final step. The problem is as follows:

Consider the function $f\left(x\right)=\text{e}^x-\dfrac{x^2}{2}$. Let $a$ and $b$ be real numbers such that $a\neq b$ and $f^\prime\left(a\right)=f^\prime\left(b\right)$.
Prove that $f\left(a\right)+f\left(b\right)<2$.

And here's my approach (failed):

Given that $f'(x) = \text{e}^x - x$, it implies that $\text{e}^a - a = \text{e}^b - b$. Let's set $a = \ln m$ and $b = \ln n$. By symmetry, we can assume $b < 0 < a$, which leads to $0 < n < 1 < m$. This results in $m - \ln m = n - \ln n$.

Hence, $m - n = \ln m - \ln n = \ln \dfrac{m}{n}$. Let's denote $t = \dfrac{m}{n}$, then $t > 1$ and $m = nt$. Consequently, $nt - n = \ln t$, which leads to $n = \dfrac{\ln t}{t - 1}$ and $m = nt = \dfrac{t \ln t}{t - 1}$.

Thus, $f(a) + f(b) = \text{e}^a + \text{e}^b - \dfrac{a^2}{2} - \dfrac{b^2}{2}$

$= m + n - \dfrac{1}{2}\left(\ln^2m + \ln^2n\right)$

$= \dfrac{t + 1}{t - 1} \cdot \ln t - \dfrac{1}{2}\left[\ln^2\left(\dfrac{t \ln t}{t - 1}\right) + \ln^2\left(\dfrac{\ln t}{t - 1}\right)\right]$

Now, the task is to prove that this function of $t$, denoted as $g(t)$, is always less than 2.

However, I am unable to prove this. Upon graphing, I found that the function is monotonically increasing on the interval $(0, 1)$, monotonically decreasing on $(1, +\infty)$, and has a removable discontinuity at the point $(1, 2)$, with $\lim\limits_{t \to 1} g(t) = 2$.

Here's what GeoGebra gave me on this function

I would greatly appreciate any guidance or alternative approaches. This problem has been a challenging puzzle for me, and as a beginner with a strong interest in mathematics, I am eager to learn.

Answer 1

As we already know $a > 0 > b$. And if $a \to 0$ we must have $b \to 0$.

If we treat $b$ as a function of $a$, satisfying

$$e^{a} - a = e^b - b$$

then $\dfrac{d b}{d a} = \dfrac{e^a - 1}{e^b - 1}$. Note $e^b - 1 < 0$.

Now let $h(a) = f(a) + f(b) - 2$, then

$$\dfrac{dh}{da} = f'(a) + f'(b) \dfrac{d b}{d a} = f'(a) (1 + \dfrac{db}{da})$$

because $f'(a) = f'(b)$.

Now we only have to deal with $(1 + \dfrac{d b}{da}) = \dfrac{e^a + e^b - 2}{e^b - 1}$

We first prove $|b|>|a|$.

Because $e^a > 1 + a + \dfrac{a^2}{2}$ and $e^b < 1 + b + \dfrac{b^2}{2}$, then

$$1 + \frac{b^2}{2} > e^b - b = e^a - a > 1 + \dfrac{a^2}{2}$$

Next, we prove $e^a + e^b - 2 > 0$.

Let $g(a) = e^a + e^b - 2$, then

$$g'(a) = e^a + e^b \frac{d b}{da} = e^a + e^b \frac{e^a - 1}{e^b - 1} = \frac{2e^{a+b}-e^a - e^b}{e^b - 1} = \frac{e^{a+b}}{e^{b}-1}(2 - e^{-a} - e^{-b})$$

Because $|b|>|a|$, $e^{-b} > e^a$, thus $2 - e^{-a} - e^{-b}<0$. Thus $g'(a)>0$.

Then $g(a) > g(0) = 0$, which means $1 + \dfrac{d b}{d a} < 0$.

Therefore $h'(a) < 0$ and $h(a) < h(0) = 0$. Done.

Answer 2

It suffices to prove the following result.

Fact 1. Let $a < 0 < b$ with $\mathrm{e}^a - a = \mathrm{e}^b - b$. Then $\mathrm{e}^a - \frac{a^2}{2} + \mathrm{e}^b - \frac{b^2}{2} < 2$.

Proof of Fact 1.

We split into two cases.

Case 1. $a < -\ln 2$

Using $\mathrm{e}^b = \mathrm{e}^a - a + b$, we have \begin{align*} \mathrm{e}^a - \frac{a^2}{2} + \mathrm{e}^b - \frac{b^2}{2} - 2 &= 2\mathrm{e}^a -a + b - \frac{a^2}{2} - \frac{b^2}{2} - 2\\ &= 2\mathrm{e}^a - 1 - \frac12(b - 1)^2 - \frac12(a + 1)^2\\ &< 0. \end{align*}

Case 2. $-\ln 2 \le a < 0$

Using $\mathrm{e}^b = \mathrm{e}^a - a + b$, we have \begin{align*} \mathrm{e}^a - \frac{a^2}{2} + \mathrm{e}^b - \frac{b^2}{2} - 2 &= 2\mathrm{e}^a -a + b - \frac{a^2}{2} - \frac{b^2}{2} - 2\\ &< 2\cdot \left(1 + a + \frac12 a^2\right) - a + b - \frac{a^2}{2} - \frac{b^2}{2} - 2\\ &= \frac12(a + 1)^2 - \frac12(b - 1)^2\\ &= \frac12(a + 2 - b)(a + b)\\ &= \frac12(2a + 2 - (a + b))(a + b) \tag{1} \end{align*} where we use $\mathrm{e}^a < 1 + a + \frac12a^2$ for all $a < 0$.
(Note: Let $h(a) = 1 + a + \frac12a^2 - \mathrm{e}^a$. We have $h'(a) = 1 + a - \mathrm{e}^a < 0$ and $h(0) = 0$.)

Since $2a + 2 \ge -2\ln 2 + 2 > 0$, from (1), it suffices to prove that $a + b < 0$.

Note that $x\mapsto \mathrm{e}^x - x$ is strictly decreasing on $(-\infty, 0)$, and strictly increasing on $(0, \infty)$.

We claim that $\mathrm{e}^a - a < \mathrm{e}^{-a} - (-a)$ for all $a < 0$. Indeed, let $g(a) = \mathrm{e}^{-a} + 2a - \mathrm{e}^a$; By AM-GM, we have $g'(a) = 2 -(\mathrm{e}^{-a} + \mathrm{e}^a) < 2 - 2\sqrt{\mathrm{e}^{-a} \cdot \mathrm{e}^a} = 0$ on $a < 0$. Also, $g(0) = 0$. The claim is proved.

Thus, we have $\mathrm{e}^b - b = \mathrm{e}^a - a < \mathrm{e}^{-a} - (-a)$ which results in $b < -a$. Thus, $a + b < 0$.

We are done.