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please help me with this problem:

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.

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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 $4\sqrt[4]{\frac{1}{3^3}}$.

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    $\begingroup$ Nice. But not obvious! $\endgroup$ – Simon S Nov 18 '14 at 18:16
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    $\begingroup$ 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 $x + y^3$. $\endgroup$ – tosi3k Nov 18 '14 at 18:21
  • $\begingroup$ That's helpful. In so doing we get a concrete number for the lower bound. Add that to the tool kit. $\endgroup$ – Simon S Nov 18 '14 at 18:26
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    $\begingroup$ 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 $a_1 = a_2 = ... = a_n$ $\endgroup$ – tosi3k Nov 18 '14 at 18:57
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    $\begingroup$ 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 $y^3 = \frac{y^{-1}}{3}$. $\endgroup$ – tosi3k Nov 18 '14 at 20:43
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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:

http://calcpad.net/Calculator

You have to type the following code:

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

Almost all numerical methods can find a local minimum in a certain interval. That is why, it is good to plot the function first to see approximately where to search. Press "Print preview" and you will get the answer on the screen like:

Plot of f(x) = x + 1/x^3

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