Determine Minimum Value.

Question

Find the minimum value of $$\frac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}$$ for $x>0$.

When $x=1$, $$\frac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}=6$$

I tried to plot some points on a graph and I observed that the minimum value is $6$.

Any hints would be sufficient. Thanks

I think differentiation would be really complicated

Answer 1

$$\dfrac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}=\dfrac{(x+1/x)^6-(x^3+1/x^3)^2}{(x+1/x)^3+(x^3+1/x^3)}=(x+1/x)^3-(x^3+1/x^3)$$ Therefore your function reduces to $3(x+1/x)$ which has its minimum when $x=1.$

Answer 2

A nice trick:

As $x>0$, $$f(x)=\frac{\left(x+\cfrac1x\right)^6-\left(x^6+\cfrac1{x^6}\right)-2}{\left(x+\cfrac1x\right)^3+\left(x^3+\cfrac1{x^3}\right)}=\frac{6x^4+15x^2+18+\cfrac{15}{x^2}+\cfrac6{x^4}}{2x^3+3x+\cfrac3x+\cfrac2{x^3}}$$ giving $$f(x)=\frac{6x^8+15x^6+18x^4+15x^2+6}{2x^7+3x^5+3x^3+2x}=\frac{6x^8+9x^6+9x^4+6x^2}{2x^7+3x^5+3x^3+2x}+\frac{6x^6+9x^4+9x^2+6}{2x^7+3x^5+3x^3+2x}$$ which boils down to $$f(x)=3x+\frac3x.$$

Answer 3

Hint-$$\frac{(x+\frac1x)^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+(x^3+\frac{1}{x^3})} = \frac{(x+\frac{1}{x})^6-(x^3+\frac{1}{x^3})^2}{(x+\frac{1}{x})^3+(x^3+\frac{1}{x^3})}=(x+\frac{1}{x})^3-(x^3+\frac{1}{x^3})=3(x+1/x)$$