AM-GM for Juniors

Algebra Level 4

Minimize $\large x^2+8x+\dfrac{64}{x^3}$ where $x$ is positive real.

24 26 28 30

2 solutions

Dexter Woo Teng Koon
Aug 23, 2017

$x^2+8x+\dfrac{64}{x^3}$

$=x^2+2x+2x+2x+2x+\dfrac{32}{x^3}+\dfrac{32}{x^3}$

$\ge 7\sqrt[7]{\left(x^2\right)\left(2x\right)\left(2x\right)\left(2x\right)\left(2x\right)\left(\dfrac{32}{x^3}\right)\left(\dfrac{32}{x^3}\right)}$

$=28$

Equality holds iff $x=2$

Marco Brezzi
Aug 20, 2017

$f(x)=x^2+8x+\dfrac{64}{x^3} \phantom{cccccc} x>0$

Taking the first derivative and setting it equal to $0$ to find the critical points

$f'(x)=2x+8-\dfrac{192}{x^4}=0 \iff x^5+4x^4-96=0$

$\Longrightarrow (x-2)(x^4+6x^3+12x^2+24x+48)=0$

Since the second parenthesis is always positive for positive $x$ , the only critical point is $x=2$

$f''(2)=\left. 2+\dfrac{768}{x^5}\right|_2>0$

So it is a point of minimum, thus

$f(2)=2^2+8\cdot 2+\dfrac{64}{2^3}=\boxed{28}$

@Marco Brezzi Can you please explain why AM-GM is not working for this f(x)

Sabhrant Sachan - 3 years, 9 months ago

Sure. If you were trying to use AM-GM like this:

$x^2+8x+\dfrac{64}{x^3}\geq 3\sqrt[3]{x^2\cdot 8x\cdot\dfrac{64}{x^3}}=24$

You can't now say that $24$ is the minimum, since equality holds only when

$x^2=8x=\dfrac{64}{x^3}$

Which can't happen

Marco Brezzi - 3 years, 9 months ago

AM-GM works :)

Dexter Woo Teng Koon - 3 years, 9 months ago