# Inequality using Calculus

#### jacks

##### Well-known member
Prove that $\displaystyle \sin x+2x \geq \frac{3x(x+1)}{\pi}\forall x\in \left[0,\frac{\pi}{2}\right]$

#### tkhunny

##### Well-known member
MHB Math Helper
I'm tempted to "use calculus" to produce the Taylor Series for f(x) = sin(x) and then observe that:

$$\sin(x) < x - \frac{x^{3}}{6}$$

#### Ackbach

##### Indicium Physicus
Staff member
I think I would approach it this way: minimize the function
$$f(x)=\sin(x)+2x-\frac{3x(x+1)}{\pi}$$
on the interval $[0,\pi/2]$ using the standard calculus techniques. There's one critical point near $0.88$, and then you have the endpoints. The left endpoint, I think, will end up being the smallest point on the graph. The second derivative might come in handy.

#### awkward

##### Member
Here is a slightly different approach.

Using f(x) as defined in post #3, show that $$f''(x) < 0$$ for all x, so f is concave. By computation, show that $$f(0) = 0$$ and $$f(\pi / 2) > 0$$. This is enough to conclude that $$f(x) > 0$$ on $$[0 , \pi/2]$$.