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.

  • $\begingroup$ $(2x)^{1/x} + (2y)^{1/y} + \frac{1}{2}x(1-x)(2x-1)^2 \le 2$ 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, 2020 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, 2020 at 15:19
  • $\begingroup$ @c-love-garlic Thanks. $\endgroup$
    – River Li
    Aug 14, 2020 at 15:30

2 Answers 2


$$(2x)^{\frac 1 x}=\frac{1}{\left( \frac{1}{2x}\right) ^{\frac{1}{x}}}\leq \frac{1}{1+\frac{1}{x}\left( \frac{1}{2x}-1\right)}=\frac{2x^2}{2x^2-2x+1}$$ 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 $=x^2+y^2$)

  • $\begingroup$ You should select this answer, which is more elementary and generalisable +1. Further there is an error in my post. $\endgroup$
    – Macavity
    Aug 14, 2020 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. $\endgroup$ Aug 14, 2020 at 9:00

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 :


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 $(1)$

  • $\begingroup$ Can someone tells me if the majorization is correct ? Thanks ! $\endgroup$
    – Erik Satie
    Aug 15, 2020 at 8:13
  • $\begingroup$ @RiverLi Can you confirm ?Good day ! $\endgroup$
    – Erik Satie
    Aug 15, 2020 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, 2020 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, 2020 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, 2020 at 16:42

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