minimum value of $\sum (x+1/x)^{10}$ expression

Question

If $x,y,z>0$ . Then prove that $\displaystyle \bigg(x+\frac{1}{x}\bigg)^{10}+\bigg(y+\frac{1}{y}\bigg)^{10}+\bigg(z+\frac{1}{z}\bigg)^{10}\geq \frac{10^{10}}{3^{9}}.$

What i try

Let $\displaystyle f(x)=\bigg(x+\frac{1}{x}\bigg)^{10}.$ Then $\displaystyle f'(x)=10\bigg(x+\frac{1}{x}\bigg)^{9}\bigg(1-\frac{1}{x^2}\bigg)>0$ fir all $x$

So $$\frac{f(x)+f(y)+f(z)}{3}\geq f\bigg(\frac{x+y+z}{3}\bigg)$$

How do i find minimum of $f((x+y+z)/3)$ help me please

Answer 1

Since $x+{1\over x}\geq 2$ with equality iff $x=1$ the actual minimum is $$3\cdot 2^{10}$$ which is smaller than one you offer.

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And if a constraint $x+y+z=1$ would be added, then you would get what you are suggesting.

Answer 2

If it means that $x+y+z=1$ so by Jensen for $f(x)=x^{10}$ and by C-S we obtain: $$\sum_{cyc}\left(x+\frac{1}{x}\right)^{10}\geq\frac{1}{3^9}\left(\sum_{cyc}\left(x+\frac{1}{x}\right)\right)^{10}=\frac{1}{3^9}\left(1+(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\right)^{10}\geq$$ $$\geq\frac{1}{3^9}\left(1+9\right)^{10}=\frac{10^{10}}{3^9}.$$