$f(x)=(x-1)\log {x}$, and $f(x_1)=f(x_2)=m, 0<x_1<x_2$. Show that $\frac{9}{5}+\log{(1+m)}<x_1+x_2<2+\frac{m}{2}$.

If we apply Hermite-Hadamard inequality, it's easy to show $2<x_1+x_2$, but it's not strong enough. I also tried to replace $m$ with $f(x_1)$, then it became super complicated. There should be some easy way to simplify $x_1, x_2$

  • $\begingroup$ en.wikipedia.org/wiki/Logarithmic_mean $\endgroup$ – function sug Jun 21 '18 at 10:03
  • $\begingroup$ What is $m$? Is for instance $m=0$ allowed? $\endgroup$ – dan_fulea Jun 21 '18 at 22:31
  • $\begingroup$ @dan_fulea m can't be 0. since 0 is the minimum value of f(x), then m=0 implies x1=x2 $\endgroup$ – Takanashi Jun 22 '18 at 0:09
  • $\begingroup$ Seeing from the graph (where $a_1=x_1-1$ and $a_2=x_2-1$), both the upper and the lower bound are quite tight. $\endgroup$ – Saad Jun 22 '18 at 9:49
  • $\begingroup$ Well, we have no domain of definition of $f$. For instance, somebody can think of "my" $f:(0,\infty)\to\Bbb R$. We have no domain where $m$ is allowed to take values. Computing "my" $f$ at $1/e$ we get a negative value, the minimum argument makes only sense when we know where $f$ is defined, but even so, why not explicitly write down the conditions?! Please complete the details. $\endgroup$ – dan_fulea Jun 22 '18 at 9:51

The upper bound of $2+\frac m2$ is proven.

For $0<x_1<1$ and $x_2>1$, the equations $$(x_1-1)\ln x_1=m\implies x_1=1+\frac m{\ln x_1}\\(x_2-1)\ln x_2=m\implies x_2=1+\frac m{\ln x_2}$$ can be added to give $x_1+x_2=2+m\left(\frac1{\ln x_1}+\frac1{\ln x_2}\right)$ and for the upper bound of $2+\frac m2$ it suffices to prove that $$\frac1{\ln x_1}+\frac1{\ln x_2}<\frac12\impliedby x_2>\exp\left(\frac{2\ln x_1}{\ln x_1-2}\right).$$ The smallest value of $x_2$ will yield $\frac1{\ln x_1}+\frac1{\ln x_2}$ closest to $\frac12$, so in the worst case, we suppose equality of the above. Substituting this equation into the one for $x_2$, we get $$(x_2-1)\ln x_2>\left[\exp\left(\frac{2\ln x_1}{\ln x_1-2}\right)-1\right]\frac{2\ln x_1}{\ln x_1-2}$$ and in turn it suffices to show that $$\left[\exp\left(\frac{2\ln x_1}{\ln x_1-2}\right)-1\right]\frac{2\ln x_1}{\ln x_1-2}\ne (x_1-1)\ln x_1$$ for $0<x_1<1$ as we want to prove that the LHS is always smaller than the RHS (e.g. at $x_1=\frac12$ this is true so if there is an interval where the LHS is greater than there must be a point where the two curves meet due to continuity).

Thus, suppose that they are equal. Rearranging gives $$f(x_1)=2x_1^{\frac2{\ln x_1-2}}+2x_1-4-(x_1-1)\ln x_1=0$$ Now $$f'(x_1)=-\frac8{x_1(\ln x_1-2)^2}x_1^{\frac2{\ln x_1-2}}+\frac1{x_1}-\ln x_1+1$$ and letting $u=2-\ln x_1$ results in $$f'(u)=\frac{1}{\exp\left(2-u\right)}-\frac{8}{u^2}\exp\left(u-\frac{4}{u}\right)+u-1$$ for $u\in(2,\infty)$. This is now computationally viable and W|A shows that indeed $f'(u)>0$ for $u\in(2,\infty)$, and $f'(u)=0$ only at $u=2$. Hence there does not exist $x_1$ such that $f(x_1)=0$, so the upper bound holds for all $x_1,x_2$ in their respective domains.


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