Prove that $\frac{\sin t\pi}\pi\ge t-2t^3+t^4$

Question

Let $t$ a real number such that $t\in[0,1]$, prove that: $$\frac{\sin t\pi}\pi\ge t-2t^3+t^4$$ I considered the function obtained by subtracting the RHS from the LHS. When I differentiate it, it gIves something not obvious to study.

Answer 1

We need to show that

$$f(t)=\frac{\sin(t\pi)}{\pi}- t+2t^3-t^4\ge 0 \quad t\in[0,1]$$

let $t=x+\frac12$ with $x\in[-\frac12,\frac12]$ thus we need to show that

$$f(x)=\frac{\cos(\pi x)}{\pi}-x^4+\frac32x^2-\frac5{16}\ge 0$$

since $f(x)$ is even we can consider $x\in[0,\frac12]$.

Note also that

• $f(0)=\frac1{\pi}-\frac5{16}>0$
• $f(1/2)=0$

then it suffices to show that $f(x)$ is monotonic on the inteval $x\in[0,\frac12]$.

Then consider

$$g(x)=f'(x)=-\sin \pi x+3x-4x^3$$

with

• $g(0)=0$
• $g(1/4)<0$
• $g(1/2)=0$

then it suffices to show that $g(x)\le 0$ on the interval $x\in[0,\frac12]$ which is true if we show that $g(x)$ has exactly one minimum on that interval.

Then consider

$$h(x)=g'(x)=-\pi\cos \pi x+3-12x^2$$

with

• $h(0)=3-\pi<0$
• $h(1/2)=0$

then it suffices to show that $h(x)=0$ has exactly one solution on the interval $x\in(0,\frac12)$ and hence exactly one maximum on that interval.

Then consider

$$l(x)=h'(x)=\pi^2\sin \pi x-24x$$

with

• $l(0)=0$
• $l(1/2)=\pi^2-12<0$

then it suffices to show that $l(x)=0$ has exactly one solution on the interval $x\in(0,\frac12)$ and hence exactly one maximum on that interval.

Then consider

$$m(x)=l'(x)=\pi^3\cos \pi x-24$$

with

• $m(0)=\pi^3-24>0$
• $m(1/2)=-24<0$

then it suffices to show that $m(x)=0$ has exactly one solution on the interval $x\in(0,\frac12)$.

Then consider

$$n(x)=m'(x)=-\pi^4\sin \pi x$$

which is negative on the interval $x\in(0,\frac12)$ and therefore on the interval $x\in(0,\frac12)$

• $m(x)$ has exactly one root
• $l(x)$ has exactly one maximum and one root
• $h(x)$ has exactly one maximum and one root
• $g(x)$ has exactly one minimum and is negative
• $f(x)$ is monotonic and positive

thus

$$f(x)=\frac{\cos(\pi x)}{\pi}-x^4+\frac32x^2-\frac5{16}\ge 0\quad \square$$