A simpler way to prove that $\frac 32-x-\frac{1}{x+1}$ is negative for $x\geq1$, instead of computing derivative? I have this function
$$f(x)=\frac 32-x-\frac{1}{x+1}$$
I want to prove that $f(x)$ is negative for $x\geq1$. We can easily prove this by calculating the first derivative. Is there a simpler way to prove that? Instead of taking derivative?
 A: Just note that, for $x \ge 1$,
$$-x - \frac{1}{x + 1} \le -\frac{3}{2} \quad \Leftrightarrow \quad 0 \le \frac{2x^2-x-1}{2(x + 1)}\quad \Leftrightarrow \quad 0 \le 2x^2 - x -1 = (x - 1)(2x+1)$$
and the last inequality clearly holds.
A: \begin{align} \left(-x-\frac{1}{x+1}\leq -\frac{3}{2}\right)\quad \text{ and }\quad  x\geq1\\
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\iff\quad \left(x+\frac{1}{x+1}\geq \frac{3}{2}\right)\quad \text{ and }\quad  x\geq1  \\
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\iff \quad \left(x(x+1) + 1 \geq \frac{3}{2}(x+1)\right)\quad \text{ and }\quad  x\geq1\\
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\iff \quad \left(x^2 - \frac{1}{2}x - \frac{1}{2} \geq 0\right)\quad \text{ and }\quad  x\geq1\\
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\iff\quad \left(\left(x-\frac{1}{4}\right)^2 \geq \frac{9}{16}\right)\quad \text{ and }\quad  x\geq1\\
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\iff\quad \left(x - \frac{1}{4} \geq \frac{3}{4}\quad \text{ or }\quad x - \frac{1}{4} \leq -\frac{3}{4}\right)\quad \text{ and }\quad  x\geq1\\
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\iff\quad \left(x\geq1\quad \text{ or }\quad x\leq -\frac{1}{2}\right)\quad \text{ and }\quad x\geq1\\
\\
\iff\quad x\geq1\\
\end{align}
