Minima value of a function

Question

Let $f$ be a function $f: \mathbb R \to \mathbb R$. $f$ defined by $$f(x) = |\cos(x)|^3 + |\sin(x)|^3$$

How can I show $f$ has a minimum value, positive and smaller than $1$?

Answer 1

$$f(x) = |\cos x|^3+|\sin x|^3$$

The triangle inequality tells us that $|a|+|b|\geq |a+b|$

So $$f(x) = |\cos x|^3+|\sin x|^3 = |\cos^3x|+|\sin ^3x|\geq|\sin ^3x+\cos^3x|$$

Consider $g(x) = |\sin ^3x+\cos^3x|$

Clearly $g(x) \geq 0$

$\sin ^3 x+\cos^3x=0\implies \tan^3x=-1\implies \tan x=-1\implies x=-\frac{\pi}{4}$

So the minimum of $g(x)$ is $0$.

Therefore we have $f(x)\geq 0$. If we can show that $f(x) \neq 0$, and $\exists a\quad\text{such that}\quad 0<f(a)<1$, then we are done.

$0<f(\frac{\pi}{4}) =\frac{\sqrt{2}}{2}<1$

Suppose for contradiction that $\exists b\quad\text{such that} \quad f(b)=0$

$|\sin b |^3=-|\cos b|^3\implies |\sin b|=-|\cos b|\implies \sin^2 b-\cos ^2b =0$

But we have the identity which holds $\forall \theta\in\mathbb{R}:\quad\sin^2\theta+\cos^2\theta=1$

Taking $\theta =b$, and then summing the two equations we get:

$2\sin^2b = 1\implies \sin b=\pm\frac{1}{\sqrt{2}}$. Substituting in the resulting values of $b$, we get $f(b)\neq 0$ which is a contradiction. Therefore we have proved that $f(x)>0$ and $\exists a\in\mathbb{R}: 0<f(a)<1$ - i.e. the minimum of $f(x)$ is positive and less than $1$.

Answer 2

Hint: If $a,b \geq 0$ then

$$\left(\dfrac{a^3+b^3}{2}\right)^{1/3}\geq \left(\dfrac{a^2+b^2}{2}\right)^{1/2}$$

Answer 3

Hint:

You can suppose $0\le x\le\pi/2$. The problem comes down to finding the minimum of $$\cos^3x+\sin^3x=(\cos x+\sin x)(\cos^2x-\sin x\cos x+\sin^2x)=\sqrt2\sin\bigl(x+\tfrac\pi4\bigr)\bigl(1-\tfrac12\sin2x\bigr).$$ On the interval $(0,\pi/2)$, $\sin\bigl(x+\tfrac\pi4\bigr)\ge\frac{\sqrt2}2$ and $\;1-\frac12\sin2x\ge\frac12$.