# Finding the minimum value of a function without using Calculus

Find the minimum value the function $$f(x) = x^4 + \frac{1}{x^2}$$ when $$x \in \Bbb R^*$$

My attempt:

Finding the minimum value of this function using calculus is a piece of cake. But since this question appeared on my test when Calculus was not taught to me, there must definitely be a way (Probably by purely using Algebra) to find the minimum value of this function without using Calculus which I am unaware of.

I tried making perfect squares but that got me nowhere. Maybe, I wasn't making the perfect, perfect square :)

Any help would be appreciated.

Use AM-GM: $$x^4+\frac1{x^2}=x^4+\frac1{2x^2}+\frac1{2x^2}\ge 3\sqrt[3]{\frac1{4}},$$ equality occurs when $$x^4=\frac1{2x^2}=\frac1{2x^2} \Rightarrow x=\pm\frac1{\sqrt[6]{2}}$$.

Hint:

As $$x^4,1/x^2>0$$ using Weighted Form of Arithmetic Mean-Geometric Mean Inequality

$$\dfrac{ax^4+bx^{-2}}{a+b}\ge\sqrt[a+b]{x^{4a-2b}}$$

Set $$4a-2b=0\iff b=2a$$

• @farruhota, Sorry for the typo. I was too much engrossed with $b=2a$ – lab bhattacharjee Mar 9 at 17:48

First, the minimum value of $$x$$ is $$b$$ such that $$f(x) - b$$ has a (double) root. (That is, the amount you must shift the graph of $$f$$ down so that it meets the $$x$$-axis once.) So we are looking for a $$b$$ so that $$f(x) - b = x^4 + \frac{1}{x^2} - b = 0$$ has a (double) root. Observe $$x^4 + \frac{1}{x^2} - b = \frac{x^6 - b x^2 + 1}{x^2}$$ has a root exactly when its numerator does. So now we just need to know when that cubic in $$x^2$$ has a double root.

The discriminant of $$(x^2)^3 - b(x^2) + 1$$ is $$-4(-b)^3 - 27 \cdot (1)^2 = 4b^3 - 27 \text{.}$$ The discriminant is zero if and only if the polynomial has a double root. Taking $$b = \sqrt[3]{27/4} = \frac{3}{2^{2/3}}$$ is the only choice that makes the discriminant zero, so the minimum value of $$f$$ is this $$b$$.