# Inequality $\arctan x ≥ x-x^3/3$

Can you help me prove $\arctan x ≥ x-x^3/3$? I have thought of taylor but I have not come up with a solution.

• what is $x$ here a real variable? Jan 12, 2015 at 19:24
• this is not true for all real $x$! Jan 12, 2015 at 19:26
• x is a real variable have you found anything? Jan 12, 2015 at 19:28
• arctan(-1) < -1 +1/3 Jan 12, 2015 at 19:35

$$1 - t^4 \le 1$$ so $$1-t^2 \le \frac1{1+t^2}$$ now integrate $$\int_0^x (1-t^2)dt \le \int_0^x \frac{dx}{1+t^2}$$ i.e. $$x-\frac{x^3}3 \le \arctan x$$

• oooo that is the correct one i think Jan 12, 2015 at 19:39
• @user2345215 what do you think? Jan 12, 2015 at 19:44
• @Argyris It's basically the same solution as mine, but written in a trickier way in order to look cool. Jan 12, 2015 at 19:44
• yes i realised it just now because it is from 0 to x thanks you have helped me a lot you know your math Jan 12, 2015 at 19:45

Have you tried derivatives? $$(\arctan x-x+x^3/3)'=\frac 1{1+x^2}-1+x^2=\frac{x^4}{1+x^2}\ge0$$ So the difference is an increasing function. This fact, together the equality when $x=0$ means that $$\arctan x\ge x-x^3/3\text{ when }x\ge 0\\\arctan x\le x-x^3/3\text{ when }x\le 0$$

• thanks i thought it would be more difficult because arctan=x-x3/3+x5/5... @Dr. Sonnhard Graubner Jan 12, 2015 at 19:37

let $$f(x)=\arctan(x)-x+\frac{x^3}{3}$$ for $x\geq 0$ for $x=0$ we get $f(0)=0$ and for $x>0$ we get $$f'(x)=\frac{x^4}{1+x^2}>0$$ therefore we obtain $$f(x)\geq 0$$ for all real $x$ with $x\geq 0$.

• thanks i thought it would be more difficult because arctan=x-x3/3+x5/5... @Dr. Sonnhard Graubner Jan 12, 2015 at 19:37