Inequality

Write the equation as :

$\large x^5 +x^5 +x^5 +x^5 +x^5 +x^5 +x^5 +x^5 + x^{-4} +x^{-4} +x^{-4} +x^{-4} +x^{-4} +x^{-4} +x^{-4} +x^{-4} +x^{-4} +x^{-4}$

Then simply applying A.M.-G.M. inequality :

$A.M. \geq G.M.$

$\large \implies x^5 +x^5 +x^5 +x^5 +x^5 +x^5 +x^5 +x^5 + x^{-4} +x^{-4} +x^{-4} +x^{-4} +x^{-4} +x^{-4} +x^{-4} +x^{-4} +x^{-4} +x^{-4} \geq 18$

So the minimum value of the given expression is $\large \boxed{18}$

enjoy!

$Let\quad f\left( x \right) ={ 8x }^{ 5 }+{ 10 }^{ -4 }\\ For\quad a\quad function\quad to\quad attain\quad its\quad minima,\quad \\ It's\quad derivative\quad should\quad be\quad equal\quad to\quad zero.\\ Therefore\quad differentiating\quad f\left( x \right) ,\\ \frac { d }{ dx } f\left( x \right) =0\\ \frac { d }{ dx } { 8x }^{ 5 }+\frac { d }{ dx } { 10x }^{ -4 }=0\\ 40{ x }^{ 4 }-\frac { 40 }{ { x }^{ 5 } } =0\\ Cancelling\quad 40\quad and\quad taking\quad LCM,\\ { x }^{ 9 }-1=0\\ Or,\quad x=1.\\ Now\quad put\quad x=1\quad in\quad f\left( x \right)\quad i.e.\quad calculate\\ f\left( 1 \right)=8{ (1) }^{ 5 }+{ 10(1) }^{ -4 }\\ f\left( 1 \right)=18$

CHEERS!!!:):)

Let $f(x)=8x^5+10x^{-4}.$ Then $\frac{\text{d}}{\text{d}x}f(x)=40x^4-\frac{40}{x^5}.$ For a function to attain its minimum, its derivative is equal to $0$ , i.e. $40x^4-\frac{40}{x^5}=0.$ Since $x$ is a positive real, by solving the equation we obtain $x=1$ , hence the minimum value of $f(x)$ is $\boxed{18}$ .

The trick to solving this problem is to reformulate it in such a way so that when you apply AM-GM Inequality, the $x$ 's will cancel out. With that in mind, we can rewrite $8x^5 + 10 x^{-4} = \underbrace{2x^5 + 2x^5 + 2x^5 + 2x^5}_{\text{4 times}} + \underbrace{2x^{-4} + 2x^{-4} + 2x^{-4} + 2x^{-4} + 2x^{-4}}_{\text{5 times}}$ .

Applying AM-GM gives us:

$\displaystyle \underbrace{2x^5 + 2x^5 + 2x^5 + 2x^5}_{\text{4 times}} + \underbrace{2x^{-4} + 2x^{-4} + 2x^{-4} + 2x^{-4} + 2x^{-4}}_{\text{5 times}} \geq 9\cdot\sqrt[\large 9]{\left(2^4 x^{20}\right)\left(2^5 x^{-20}\right)} = 18$

Therefore, the minimum value of the expression is $\boxed{18}$ .