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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.

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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 $x^4=\frac1{2x^2}=\frac1{2x^2} \Rightarrow x=\pm\frac1{\sqrt[6]{2}}$.

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

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

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