9
$\begingroup$

For positive reals $x$ and $y$ such that $x+y=1$, prove that $$(2x)^{\frac 1 x}+(2y)^{\frac 1 y}\leq 2$$

I have tried using Jensen’s inequality but it won’t cover all the possible choices for $x$ and $y$ since the concavity varies. I am trying to find a neat solution so that a generalization could also be made. Thank you.

| cite | improve this question | |
$\endgroup$
  • $\begingroup$ $(2x)^{1/x} + (2y)^{1/y} + \frac{1}{2}x(1-x)(2x-1)^2 \le 2$ is also true. We may determine the best constant $k$ such that $(2x)^{1/x} + (2y)^{1/y} + k x(1-x)(2x-1)^2 \le 2$ is true. $\endgroup$ – River Li Aug 14 at 14:57
  • $\begingroup$ @RiverLi you can use my answer an strongly convex function ncbi.nlm.nih.gov/pmc/articles/PMC6244717 . $\endgroup$ – Erik Satie Aug 14 at 15:19
  • $\begingroup$ @c-love-garlic Thanks. $\endgroup$ – River Li Aug 14 at 15:30
11
$\begingroup$

$$(2x)^{\frac 1 x}=\frac{1}{\left( \frac{1}{2x}\right) ^{\frac{1}{x}}}\leq \frac{1}{1+\frac{1}{x}\left( \frac{1}{2x}-1\right)}=\frac{2x^2}{2x^2-2x+1}$$ by Bernoulli’s inequality. The same holds for $y$ and one immediately computes that if $x+y=1$, $$\frac{2x^2}{2x^2-2x+1}+\frac{2y^2}{2y^2-2y+1}=2$$ and the result follows.(just observe that the denominator are the same $=x^2+y^2$)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ You should select this answer, which is more elementary and generalisable +1. Further there is an error in my post. $\endgroup$ – Macavity Aug 14 at 7:04
  • $\begingroup$ @William Sun By derivatives we can prove that for any $x>0$ we have: $(2x)^{\frac{1}{x}}\leq\frac{1}{30}(6144x^7-20352x^6+26720x^5-16240x^4+2880x^3+1568x^2-694x+75),$ but it leads to wrong inequality. $\endgroup$ – Michael Rozenberg Aug 14 at 9:00
0
$\begingroup$

Question : Can we use Jensen's inequality here ? No .

Hint :

One can show that the function : $$f(x)=1-\frac{1}{(2a(1-x)+2bx)^{\frac{1}{a(1-x)+bx}}}$$ Is concave on $I=[0,1]$ where $1>a\geq 0.5\geq b>0$ and $a+b=1$

So we can apply Jensen's inequality with weight we get :

$$(2a)^{\frac{1}{a}}f(0)+(2b)^{\frac{1}{b}}f(1)\leq ((2a)^{\frac{1}{a}}+(2b)^{\frac{1}{b}})f\Bigg(\frac{(2b)^{\frac{1}{b}}}{(2a)^{\frac{1}{a}}+(2b)^{\frac{1}{b}}}\Bigg)$$

But :

$$\frac{(2b)^{\frac{1}{b}}}{(2a)^{\frac{1}{a}}+(2b)^{\frac{1}{b}}}\leq 0.5$$

One can show that the function $f(x)$ is decreasing on $I$ so :

$$f\Bigg(\frac{(2b)^{\frac{1}{b}}}{(2a)^{\frac{1}{a}}+(2b)^{\frac{1}{b}}}\Bigg)\geq f(0.5)=0$$

Now we can conclude !

Update :

I think we can use weighted Karamata's inequality (but I'm not sure) to get :

$$(2a)^{\frac{1}{a}}f(0)+(2b)^{\frac{1}{b}}f(1)\leq (2a)^{\frac{1}{a}}f(0.5)+(2b)^{\frac{1}{b}}f(0.5)=0$$

Since :$$0(2a)^{\frac{1}{a}}\leq 0.5(2a)^{\frac{1}{a}} \quad 0.5(2a)^{\frac{1}{a}}+0.5(2b)^{\frac{1}{b}}\geq 1(2b)^{\frac{1}{b}}+0(2a)^{\frac{1}{a}}$$

Wich is the desired inequality .

I think that the update is false but with strong convexity it seems to work (numerically) with particular value for $a,b$ .

Curious fact :

We cannot use Jensen's inequality directly but we can use it to refine the result and I found it very strange .

One can show that the function :

$$g(x)=\frac{1}{\ln\Big((2x)^{\frac{1}{x}}+(2(1-x))^{\frac{1}{1-x}}-1\Big)}$$

where $0<x<0.5$ is concave .To prove it we can use the proof of William Sun .

Now except the extrema and the equality case the equality :

$$h(x)=(2x)^{\frac{1}{x}}+(2(1-x))^{\frac{1}{1-x}}=c\quad (1)$$

Have four solutions on $I=(0,1)$

So two solutions on $0<x<0.5$ and two solutions on $0.5<x<1$ we can express it as : $$h(x)=h(y)=h(1-y)=h(1-x)$$

Where $0<x<y<0.5$

So if we apply Jensen inequality we have :

$$g(x)+g(y)=g(x)+g(1-x)=2 g(x)\leq 2g\Big(\frac{x+y}{2}\Big)$$

To conclude we can use numerical methods to solve $(1)$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Can someone tells me if the majorization is correct ? Thanks ! $\endgroup$ – Erik Satie Aug 15 at 8:13
  • $\begingroup$ @RiverLi Can you confirm ?Good day ! $\endgroup$ – Erik Satie Aug 15 at 8:42
  • $\begingroup$ From $A \le B$ and $B \ge 0$, we can not get $A \le 0$, right? $\endgroup$ – River Li Aug 15 at 11:37
  • $\begingroup$ @RiverLi If you speak about the inequality of Jensen's it's the goal to see we cannot use it here .And we cannot use Karamata's inequality furthermore . $\endgroup$ – Erik Satie Aug 15 at 16:36
  • $\begingroup$ OK. You want to prove that we can not use Jensen. But maybe we can use Jensen in other way? $\endgroup$ – River Li Aug 15 at 16:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.