An inequality on the roots of a transcendental equation 
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?
 A: 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.
