Proof of inequality $x + x^{-1} - x^r - x^{-r} \geq 0$ for $x>0$ and $0<r<1$.

Question

I am trying to prove the inequality $$x + x^{-1} - x^r - x^{-r} \geq 0$$ for $x>0$ and $0<r<1$. I believe this to be true as it was cited in a proof that I followed, but was not proved there. Desmos shows that the function $$f(x) = x + x^{-1} - x^r - x^{-r}$$ graphed for each specific $r$ in the range seems to be convex on $(0, \infty)$ however the function given by differentiating this is challenging for me to show this. The function has a single root on $(0, \infty)$, at $x=1$ graphically and lies above the real line everywhere on this interval, however, I need to prove this rigorously.

Since $x>0$, we can multiply by any power of $x$ to give a perhaps nicer function to prove it with equivalently. I made some progress with multiplying by a single factor $x$ to give $$g(x) = x^2 + 1 - x^{r+1} - x^{1-r}$$ $$g'(x) = 2x - (r+1) x^r -(1-r)x^{-r}$$ $$g''(x) = 2 - r(r+1) x^{r-1} + r(1-r)x^{-r-1}$$ however, I am unable to complete the argument, as each of the two nonzero terms in $g''$ are hard to work with. I have also tried partitioning into the two cases where $x>1$ and $x\leq1$, but have had little luck with this either. I have also tried applying some form of the AM-GM inequality to no avail. It is perhaps worth noting that the original form of this inequality was as $$\frac{x + x^{-1}}{x^r + x^{-r}} \geq 1$$ however, the rearranged form feels like it should be easier. Any suggestions and ideas are appreciated. Thanks, Will.

Answer 1

Hint

Try treating $\ x^r+x^{-r}\ $ as a function of $\ r\ $ for fixed $\ x>0\ $. What happens to the value of that function as $\ r\rightarrow1\ $, and what is the sign of $\ \frac{d}{dr}\big(x^r+x^{-r}\big)\ $?

Answer 2

You can write the first member in this way: $$x=\exp(\ln(x))\text{ since }x>0\text{ this as sense.}$$ So you have:$$\exp(\ln(x))+\exp(-\ln(x))-\exp(r\ln(x))-\exp(-r\ln(x))\geq0$$ We know that: $$\cosh(z)=\frac{e^z+e^{-z}}{2}$$ So we have $$2\cosh(\ln(x))-2\cosh(r\ln(x))\geq0$$ Divide by 2 and subtract $\cosh(r\ln(x))$ $$\cosh(\ln(x))\geq \cosh(r\ln(x))$$ We apply $\operatorname{arccosh}$. (Note that $\operatorname{arccosh}(\cosh(x))=|x|\neq x$) $$|\ln(x)|\geq |r\ln(x)|$$ $$|\ln(x)|\geq |r|\cdot|\ln(x)|$$ Since $0<r<1$ we have that $|r|=r$ $$|\ln(x)|\geq r\cdot|\ln(x)|$$ $$|\ln(x)|(r-1)\leq 0$$ Now: $0<r<1$ so $0-1<r-1<1-1\;\Rightarrow -1<r-1<0$
Since $|\ln(x)|\geq 0$ and $-1<r-1<0$ the last product is true and you proved the inequality.

Answer 3

Using Bernoulli inequality $(1 + u)^r \le 1 + ur$ for all $u > -1$ and $0 < r < 1$, we have $$x^r \le 1 + (x - 1)r.$$ and $$x^{-r} = (1/x)^r \le 1 + (1/x - 1)r.$$

Thus, $$x - x^r \ge x - [1 + (x - 1)r] = (x - 1)(1 - r)$$ and $$x^{-1} - x^{-r} \ge x^{-1} - [1 + (1/x-1)r] = (1/x - 1)(1 - r).$$

Thus, $$x + x^{-1} - x^r - x^{-r} \ge (x + 1/x - 2)(1 - r)\ge 0$$ where we use $x + 1/x \ge 2$ (by AM-GM).

We are done.