An algebra problem by Kalfin Muchtar

In your problem, what are the possible values of $x$ ? Note that as $x$ approaches $\pi/2$ from above, $\tan x$ approaches $-\infty$ , so there is no minimum. You have to specify a range for $x$ , like $0 \le x < \pi/2$ .

Also, your "solution" does not make work. Just because $\sin^3 x + \cos^3 x + \tan^3 x \ge 3 \sin^2 x$ , and the minimum value of $3 \sin^2 x$ is 0, it does not mean that the minimum value of $\sin^3 x + \cos^3 x + \tan^3 x$ is also 0.

For example, consider the follwing "solution": We want to find the minimum value of $x^2 + 1$ for $x \ge 0$ . By AM-GM, $x^2 + 1 \ge 2x$ . The minimum value of $2x$ is 0. Therefore, the minimum value of $x^2 + 1$ is 0.

However, it's clear that that the minimum value of $x^2 + 1$ is 1, not 0.

To establish that the minimum value of a function $f(x)$ is $m$ , you must show things:

(1) $f(x) \ge m$ for all possible values of $x$ , and

(2) $f(x) = m$ for some value of $x$ .

Just establishing a bound using AM-GM is not sufficient. You also have to show that the minimum you claim can also be achieved.