Is Inequality Proven Using Calculus in a Different Approach?

[juantheron]

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

 

[tkhunny]

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]

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]

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 \( [0 , \pi/2] \).