# Finding lowest possible value

Find the lowest possible value of $$x+y^3$$ where both x and y are positive and x*y=1.

I know how to solve this one using my method, but I was suggested to use geometric and arithmetic mean. I have no idea how to solve this using them.

Since $xy = 1$, observe that $x + y^3 = y^{-1} + y^3 = y^3 + \frac{y^{-1}}{3} + \frac{y^{-1}}{3} + \frac{y^{-1}}{3} \geq 4\sqrt[4]{\frac{1}{3^3}}$.

The equality holds if and only if $y^3 = \frac{y^{-1}}{3}$, that means for $y = \frac{1}{\sqrt[4]{3}} > 0$, therefore your lowest possible value of $x + y^3$ is $4\sqrt[4]{\frac{1}{3^3}}$.

• Nice. But not obvious! – Simon S Nov 18 '14 at 18:16
• The intuition is to somehow get rid of "3" in the $y^3$ through the AM-GM inequality and $xy = 1$ implies taking $x = y^{-1}$ which quickly suggests such representation of $x + y^3$. – tosi3k Nov 18 '14 at 18:21
• That's helpful. In so doing we get a concrete number for the lower bound. Add that to the tool kit. – Simon S Nov 18 '14 at 18:26
• If you have positive reals $a_1, ..., a_n$, the inequality between arithmetic and geometric mean looks as follows: $$\frac{1}{n}\sum_{j=1}^{n}a_j \geq \left(\prod_{j=1}^{n}a_j\right)^{\frac{1}{n}}$$ and the inequality becomes an equality if and only if $a_1 = a_2 = ... = a_n$ – tosi3k Nov 18 '14 at 18:57
• We have a lower bound on the minimal value of $x + y^3$ equal to a constant $4\sqrt[4]{\frac{1}{3^3}}$, right? And now we want to proof that this bound is in fact our minimal value. I meant that in our AM-GM inequality application we can attain equality to this lower bound by seeking such $y$ that $y^3 = \frac{y^{-1}}{3}$. – tosi3k Nov 18 '14 at 20:43

It is to find the minimum value of $$f(x; y) = x + y^3$$ but moving only along the curve $$x*y = 1$$ in the Oxy plane, like in the picture:

Plot for f(x; y) = x + y^3

To solve this, use the condition $$x*y = 1$$, to substitute y and express the function only in terms of x:

$$y = 1/x$$

$$f(x) = x + 1/x^3$$

Then find the minimum of the function f(x) and the corresponding x. You can do it either analyticalally or numerically:

Find the first derivative $$f'(x) = 1 - 3/x^4$$

Solve it for $$f'(x) = 0$$. The root is $$x = 3^(1/4) = 1.316$$

$$y = 1/x = 0.7599$$

Finally lowest value of $$f(x; y) = 1.316 + 0.7599^3 = 1.7548$$

To solve it numerically, you can use for example the calcpad online calculator:

f(x) = x + 1/x^3
$Plot{f(x) @ x = 0.5 : 5}$Inf{f(x) @ x = 1: 2}