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Denote the two real roots of $x-\ln x=a(a> 1)$ as $x_1,x_2$. Show that $x_1+x_2\leq a+\sqrt{a}$.

If we introduce Lambert W function, we can express the roots as $$x_1=-W_{-1}(-e^{-a}),~~~~x_2=-W(-e^{-a}).\\$$ Thus we only need to show $$f(a)=-W_{-1}(-e^{-a})-W(-e^{-a})-a-\sqrt{a}\leq 0.$$ By graphing the function, we can see $f(a)$ is a decreasing function, thus $f(a)\leq f(1)=0,$ which is what we want. But how to prove this rigorously?

mengdie1982
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1 Answers1

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Update

Remark: I came up with an ugly proof. Some proofs are not hard and thus omitted. Hint for Facts 3 and 4: Use Euler's substitution $a = \frac{14+t^2}{4t+10}$ ($t>2$) to obtain $\sqrt{4a^2+10a-14} = \frac{(t+7)(t-2)}{2t+5}$.

Let $f(x) = x - \ln x - a$. Let \begin{align} x_3 &= \frac{1+3a}{3+a} + a - \frac{1}{3}(2a+1+\sqrt{4a^2+10a-14}), \\ x_4 &= \frac{1}{3}(2a+1+\sqrt{4a^2+10a-14}). \end{align}

We first give some auxiliary results (Facts 1 through 4).

Fact 1: $f(x)$ is strictly decreasing on $(0, 1)$, and strictly increasing on $(1, \infty)$. Also, $\lim_{x\to 0} f(x) = +\infty$, $f(1) < 0$ and $\lim_{x\to +\infty} f(x) = +\infty $.

Fact 2: $f(\sqrt{a} + a - \tfrac{1}{5}) > 0$ for $a > \frac{121}{64}$.

Fact 3: $x_3 \in (0, 1)$ and $f(x_3) < 0$ for $1 < a \le \frac{121}{64}$.

Fact 4: $x_4\in (1, \infty)$ and $f(x_4) > 0$ for $1 < a \le \frac{121}{64}$.

Now we proceed. From Fact 1, we have $x_1\in (0, 1)$ and $x_2 \in (1, \infty)$.

We split into two cases:

1) $a > \frac{121}{64}$: Note that $f(\frac{1}{5}) = \frac{1}{5} - \ln \frac{1}{5} - a < 0$. From Fact 1, we have $x_1 < \frac{1}{5}$. Clearly, $\sqrt{a} + a - \tfrac{1}{5} > 1$. From Fact 1 and Fact 2, we have $x_2 < \sqrt{a} + a - \tfrac{1}{5}$. Thus, we have $x_1 + x_2 < \sqrt{a} + a$.

2) $1 < a \le \frac{121}{64}$: From Facts 1, 3 and 4, we have $x_1 < x_3$ and $x_2 < x_4$. Thus, we have $x_1 + x_2 < x_3 + x_4 = \frac{1+3a}{3+a} + a \le \sqrt{a} + a$ where we have used $\sqrt{a} \ge \frac{1+3a}{3+a}$ for $a > 1$.

We are done.

River Li
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