Prove that x |ln x | ≤ 1/e for x in (0,1), with equality

Prove that $x\bigl|\ln x\bigr| \le \dfrac{1}{e}$ for $x\in\bigl(0,1\bigr)$, with equality only at $x = e^{-1}$.

Step 1: Remove the absolute value

On $\bigl(0,1\bigr)$ we have $\ln x\lt 0$, so $\bigl|\ln x\bigr| = -\ln x$ and the claim concerns

\[ f(x) = -x\ln x, \qquad 0\lt x\lt 1. \]

Step 2: Find the maximum

Extend $f$ continuously by $f(0) = 0$ (Problem 391) and $f(1) = 0$. Then

\[ f'(x) = -\ln x-1 = 0 \;\Longleftrightarrow\; \ln x = -1 \;\Longleftrightarrow\; x = e^{-1}, \] and $f'\gt 0$ for $x\lt e^{-1}$, $f'\lt 0$ for $x\gt e^{-1}$: the function rises to a single peak and falls back, so the peak is the global maximum on $\bigl[0,1\bigr]$.

Step 3: The maximum value

\[ f\!\left(\frac1e\right) = -\frac1e\ln\frac1e = \frac1e. \]

Step 4: Conclusion — and a remark on strictness

Therefore $x\bigl|\ln x\bigr|\le\dfrac1e$ throughout $\bigl(0,1\bigr)$. The bound is attained at $x = e^{-1}$, where the two sides are equal, so the strict form $x|\ln x|\lt e^{-1}$ is false at that one point and true everywhere else on the interval. The sharp statement is the one with $\le$.

Step 5: Verification

\[x = 0.1:\ 0.2302585; \qquad x = 0.2:\ 0.3218876; \qquad x = e^{-1}:\ 0.3678794 = \tfrac1e \;\;\checkmark\]

\[x = 0.5:\ 0.3465736; \qquad x = 0.9:\ 0.0948245 \;\;\checkmark\ \text{(both below the peak)}\]

\[f''(x) = -\frac1x\lt 0 \text{ on } \bigl(0,1\bigr):\ f \text{ is concave, so the critical point is the maximum} \;\;\checkmark\]