Find the minimum value of the function a(x).

Question

$$a(x)= \sqrt{x^3}+\sqrt{x^{-3}}-4(x+\frac{1}{x})$$

One of the ways I could think of was to find out the global extreme values and proceed.But as I began doing it that it takes a lot of efforts and time. Is there a simpler way to proceed? Further please suggest better tags if needed.

Answer 1

Since $x\gt 0$, we can pick $y\gt 0$ such that $x=y^2$; then the given expression becomes $$(y + {1\over y})^3 - 3(y+{1\over y}) - 4(y+{1\over y})^2 + 8$$ That's equivalent to finding the minimum of $$t^3-4t^2-3t+8$$ for $t\ge 2$

Answer 2

Hint: It is $$\sqrt{x^3}+\frac{1}{\sqrt{x^3}}-4\left(x+\frac{1}{x}\right)\geq -10$$ and this is equivalent to $$\frac{(x-1)^2(1-7x+x^2)^2}{x^3}\geq 0$$