# Find minimum of function and minimum point

We have a function $$\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})}{(x+\frac{1}{x})^3+(x^3+\frac{1}{x^3})}$$ the task to find minimum of this function and point where is minimum.

I tried to do by mark $$t=x+\frac{1}{x}$$ and then simplify this function and after that find derivative etc., but it is such a long solution. The professor, in the class, said that there is very short way to do this exercise, maybe somebody know it?

Why is that substitution bad? We get $${t^6-(t^6-6t^4+9t^2-2)\over t^3+(t^3-3t)} = {6t^4-9t^2+2\over 2t^3-3t}$$ $$= 3t+{2\over 2t^3-3t}$$

Derivative of that is $$3-{12t^2-6\over (2t^3-3t)^2}$$

So we want to solve $$4t^6-12t^4+9t^2= 4t^2-2$$i.e.

$$4t^6-12t^4+5t^2+ 2=0$$

Write $$s=t^2$$, so we have $$4s^3-12s^2+5s+2=0$$

We see that we have zeroes in following intervals $$(-1,0)$$, $$(0,1)$$ and $$(2,3)$$. Since $$s = (x+{1\over x})^2= x^2+2+{1\over x^2} \geq 2+2 =4$$ we see that this equation has no solution.

• Sorry, I did little mistake in function. There is + in the denominator. @greedoid Feb 27, 2019 at 19:31
• This is cubic in $t^2$. I think a closed form exists for solving cubic equations. Feb 27, 2019 at 19:50
• @greedoid in your 3rd expression, in numerator maybe must be +6, not -12? Feb 27, 2019 at 20:13

The minimum does not exist.

Try $$x\rightarrow0^-$$.

For finding of a local minimum we can make the following.

Let $$x=1$$.

Thus, the expression is equal to $$\frac{31}{5}.$$

We'll prove that we got a minimal value.

Indeed, $$\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)}{\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)}-\frac{31}{5}=\frac{(x-1)^2(30x^6-2x^5+41x^4-9x^3+41x^2-2x+30)}{5x(2x^6+3x^4+3x^2+2)}\geq0$$ and we are done!

Now, let $$f(x)=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)}{\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)}.$$

Thus, $$f(-x)=-f(x),$$ which says that $$\max_{x<0}f=f(-1)=-\frac{31}{5}.$$

For $$x>0$$ we can use the following.

Firstly, by Holder $$4\left(x^3+\frac{1}{x^3}\right)=(1+1)^2\left(x^3+\frac{1}{x^3}\right)\geq\left(x+\frac{1}{x}\right)^3.$$ Also, by AM-GM $$\left(x+\frac{1}{x}\right)^9=\left(x+\frac{1}{x}\right)\left(x^2+2+\frac{1}{x^2}\right)^4=\left(x+\frac{1}{x}\right)\left(x^2-1+\frac{1}{x^2}+3\right)^4\geq$$ $$\geq\left(x+\frac{1}{x}\right)\left(4\sqrt[4]{\left(x^2-1+\frac{1}{x^2}\right)\cdot1^3}\right)^4=256\left(x+\frac{1}{x}\right)\left(x^2-1+\frac{1}{x^2}\right)=256\left(x^3+\frac{1}{x^3}\right).$$ Id est, by AM-GM again we obtain: $$\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)}{\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)}=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}+2\right)+2}{\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)}=$$ $$=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^3+\frac{1}{x^3}\right)^2+2}{\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)}=\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)+\frac{2}{\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)}=$$ $$=3\left(x+\frac{1}{x}\right)+\frac{2}{\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)}\geq3\left(x+\frac{1}{x}\right)+\frac{2}{5\left(x^3+\frac{1}{x^3}\right)}=$$ $$=\frac{1}{5}\left(30\cdot\frac{x+\frac{1}{x}}{2}+\frac{2}{x^3+\frac{1}{x^3}}\right)\geq\frac{31}{5}\sqrt[31]{\left(\frac{x+\frac{1}{x}}{2}\right)^{30}\frac{2}{x^3+\frac{1}{x^3}}}\geq$$ $$\geq\frac{31}{5}\sqrt[31]{\left(\frac{x+\frac{1}{x}}{2}\right)^{9}\frac{2}{x^3+\frac{1}{x^3}}}\geq\frac{31}{5}\sqrt[31]{1}=\frac{31}{5}.$$