# Determine the lowest value of $A$ such that $f (x) \geq20$ for all $x> 0$

Be the function $$\begin{array}{lrll}f:&(0,+\infty)&\longrightarrow&\mathbb{R}\\&x&\longmapsto&f(x)=3x^{3}+\dfrac{A}{x^{3}}\end{array}$$ where $A$ is a positive constant. Determine the lowest value of $A$ such that $f (x) \geq20$ for all $x> 0$

It seems to me that I have to limit the function to values ​​greater than or equal to 20, so, I suppose I need to optimize the function, right? that is, take first derivative and choose the appropriate minimum.

• Yes, with appropriate considerations for what happens when $x \to 0$ and $x \to \infty$ this is a very good idea. Can you do this or do you need help? Commented Aug 27, 2018 at 22:29
• I would like to give some clues maybe ... because, the analysis in the extremes of the interval I did not understand you very well. Commented Aug 27, 2018 at 22:35
• hey did you understand my solution ?? or you are looking to solve it by another method Commented Aug 27, 2018 at 23:03

You can use the A.M. G.M. method because here $x^3$ and $\frac{A}{x^3}$ are positive.

$$\frac{3x^{3}+\dfrac{A}{x^{3}}}{2}\ge (3A)^{\frac{1}{2}}$$ $$3x^{3}+\dfrac{A}{x^{3}}\ge 2(3A)^{\frac{1}{2}}$$ thus $$2(3A)^{1/2}\ge20$$ $$3A\ge 100$$ $$A\ge \frac{100}{3}$$

• What's the meaning or what is A.M.G.M. method? Commented Aug 27, 2018 at 22:53
• Arithmetic mean and the geometric mean, of course, you can solve it by differentiating the function and finding the point where $f'(x)=0$ and so on Commented Aug 27, 2018 at 22:53
• I just derived and got that $A = 3x^{6}$ but I got stuck there. Commented Aug 27, 2018 at 23:09
• hey you are going in the right way $x^3=\sqrt{A/3}$ put in the equation $f(x)=3\sqrt{A/3}+\frac{A}{\sqrt{A/3}}=2\sqrt{A3}$ also there was a typo in my solution check out the updated solution of mine Commented Aug 27, 2018 at 23:12
• Thank you very much!! excuse a question ... as weapons directly the inequality, $\frac{f (x)}{2} \geq\sqrt{3A}$ the next step if I understand Commented Aug 27, 2018 at 23:19

Using algebra.

Consider the function $$f(x)=3x^{3}+\dfrac{A}{x^{3}}$$ $$f'(x)=9 x^2-\frac{3 A}{x^4}$$ $$f''(x)=\frac{12 A}{x^5}+18 x$$ The first derivative cancels when $$9 x^2-\frac{3 A}{x^4}=0 \implies x_*=\left(\frac A 3 \right) ^\frac 16$$ For this value $$f(x_*)=2 \sqrt{3A}$$ and the second derivative test shows that $x_*$ corresponds to a minimum. So, you want $2 \sqrt{3A} \geq 20$; then $A$.