How to show $|x\log x| \le 1$ on $(0,1]$? How would I formally prove that $|x\log x| \le 1$ on $(0,1]$. I tried showing that $x\log x \le x^2 \le 1$, but even showing $\log x \le x$ I'm finding difficult. And then finding a lower bound for the absolute value is driving me nuts. My apologies if this is trivial, but I just can't quite get it. I'm not allowed to use calculus techniques such as second derivative tests and such. Any aid would be great.
 A: Since $x\log x\le0$ for $0<x\le1$, you want to prove that $x\log x\ge-1$, which means
$$
\log x\ge-\frac{1}{x}
$$
that's the same as
$$
\log\frac{1}{x}=-\log x\le \frac{1}{x}
$$
With the substitution $t=1/x$, this becomes proving that, for $t\ge1$, $\log t\le t$, or $t\le e^t$.
Now choose the proof you prefer among those at Simplest or nicest proof that $1+x \le e^x$

With a bit of calculus, consider the function $f(x)=x\log x$; then
$$
f'(x)=1+\log x
$$
which is negative for $0<x<1/e$ and positive for $x>1/e$.
Since $\lim_{x\to0}f(x)=0$, $f(1/e)=-1/e$ and $f(1)=0$ you can say even more:
$$
-\frac{1}{e}\le x\log x\le 0
$$
for $x\in(0,1]$.
A: $\log(x) \leq x$ is trivial, but does not help you because $\log(x)$ is negative in $(0,1]$.
For this solution, I assume that you know $y+1 \leq \exp(y)$ for $y \geq 0$, $-\log(x) = \log\left(\frac{1}{x}\right)$ for $x \in (0,1]$ and that the logarithm is monotone. All of these facts can be proven directly by the definition of $\exp$ and $\log$ (as the inverse of $\exp$).
Let $y \geq 0$ and observe that $y+1 \leq \exp(y)$ implies $\log(y+1) \leq y$ by monotonicity. Plugging in $y = \frac{1}{x}-1$, we get
$$-\log(x) = \log\left(\frac{1}{x}\right) \leq \frac{1}{x}-1$$
and hence $|x\log(x)| = -x\log(x) \leq 1 - x \leq 1$ for $x \in (0,1]$.
