I found this problem in my old paper :

Let $f(x)$ be a convex function on $(0,\infty)$ such that $\forall x>0$ we have $f(x)>0$ and $n\geq 3$ a natural number then we have : $$\Big(f(1)^{f(1)}f(2)^{f(2)}\cdots f(n)^{f(n)}\Big)^{\frac{1}{f(1)+f(2)+\cdots + f(n)}}+\Big(f(1)f(2)\cdots f(n)\Big)^{\frac{1}{n}}\leq f(1)+f(n) $$

I try to use Jensen's inequality we have :

$$\ln\Big( \Big(f(1)^{f(1)}f(2)^{f(2)}\cdots f(n)^{f(n)}\Big)^{\frac{1}{f(1)+f(2)+\cdots + f(n)}} \Big)\leq \ln\Big(\frac{f^2(1)+f^2(2)+\cdots+f^2(n)}{f(1)+f(2)+\cdots+f(n)}\Big)$$

Remains to show this :

$$\ln\Big(\frac{f^2(1)+f^2(2)+\cdots+f^2(n)}{f(1)+f(2)+\cdots+f(n)}\Big)\leq \ln\Big(f(1)+f(n)-\Big(f(1)f(2)\cdots f(n)\Big)^{\frac{1}{n}}\Big)$$

This last inequality is true for $f(x)=e^x$ but certainly not for $f(x)=x$

Furthermore this result recall me the Mercer's inequality (see here)

Finally If the function $f(x)$ is concave and positive the inequality of the beginning is reversed .

I think it's too hard for an maths competition but you can use the tools you want .

I prefer hints as answer.

Thanks a lot for sharing your time and knowledge .

Edit :

Ooops I don't mentionned that the case $n=2$ is special it correspond to this

New bound for Am-Gm of 2 variables

Second edit :

I disturbed me a little bit but I think we can add the following constraint. I add the fact that $\ln(f(x))$ must be concave on $(0,\infty)$.If it's doesn't work too can someone prove the inequality for $f(x)=x\alpha$ where $\alpha>0$ ?

Thanks again !

  • 4
    $\begingroup$ If I calculated correctly then your inequality is wrong for $f(x)=x$ and $n=2$. The LHS is $3.001614614341294 > f(1) + f(2) = 3$. $\endgroup$ – Martin R Oct 24 '19 at 14:05
  • $\begingroup$ Posted also on MathOverflow: An olympiad-like inequality. $\endgroup$ – Martin Sleziak Oct 25 '19 at 6:07
  • $\begingroup$ why is there a bounty on this? did martin R disprove it or not? $\endgroup$ – mathworker21 Oct 31 '19 at 17:18
  • $\begingroup$ @mathworker21: Constraint is $n\ge3$ which was edited after Martin's comment. $\endgroup$ – TheSimpliFire Oct 31 '19 at 19:48
  • 1
    $\begingroup$ Thanks @GerryMyerson for spotting that. The MO post has undergone deletion, and resurrection some hours ago, and most importantly, received an answer. $\endgroup$ – Hanno Nov 2 '19 at 22:13

I think that the inequality is not true for $n=3$ and $$f(x) = \mathrm{e}^{ax^2 + bx + c}$$ where \begin{align} a &= -\frac{1}{2}\ln 10 + \frac{1}{2}\ln 800 - \ln \frac{3}{20} \approx 4.088133303, \\ b &= \frac{5}{2}\ln 10 - \frac{3}{2}\ln 800 + 4 \ln \frac{3}{20} \approx -11.85893480, \\ c &= -3\ln 10 + \ln 800 - 3\ln \frac{3}{20} \approx 5.468216404. \end{align}


First, we have $f''(x) = \mathrm{e}^{ax^2 + bx + c}(4a^2x^2 + 4abx + b^2 + 2a)$. It is easy to prove that $f''(x) > 0$ for $x > 0$. Thus, $f(x)$ is a convex function on $(0, \infty)$. Also $f(x) > 0$ is obvious.

Second, we have $f(1) = \frac{1}{10}, \ f(2) = \frac{3}{20}, \ f(3) = 800$. For convenience, denote $A = f(1), \ B = f(2), \ C = f(3)$. By using Maple software, it is easy to check that \begin{align} (A+B+C) \ln (A + C - (ABC)^{1/3}) - \Big(A\ln A + B\ln B + C\ln C\Big) \approx -0.007135. \end{align} This disproves the inequality.

  • 1
    $\begingroup$ Will the op move the goal post by excluding $n=3$ again? $\endgroup$ – WE Tutorial School Nov 3 '19 at 17:41
  • 1
    $\begingroup$ Or add some constraints on $f(1), f(2), \cdots, f(n)$. $\endgroup$ – River Li Nov 4 '19 at 1:49
  • 1
    $\begingroup$ @WETutorialSchool: And indeed – the question has changed again :) $\endgroup$ – Martin R Nov 5 '19 at 10:32
  • 1
    $\begingroup$ @Konstantin First, Maple software provide high precision results. Second, if we fix $A, B$ and let $C\to \infty$, Maple software gives $\lim_{C\to \infty} (A+B+C)\ln (A+C - (ABC)^{1/3}) - (A\ln A + B\ln B + C\ln C) = -\infty$. With this in mind, we can actually disprove the inequality analytically. $\endgroup$ – River Li Nov 6 '19 at 13:17
  • 1
    $\begingroup$ @Konstantin Yes. I do not know what if $\ln f(x)$ is concave. By the way, it is not hard to prove that the inequality is true if $f(x) = \alpha x$ for $\alpha > 0$. $\endgroup$ – River Li Nov 6 '19 at 16:19

Hint: Normalize both sides by $\sum_i f(i)$ and take a look at case $n=3$:

Geometry of the case n=3

Red axis corresponds to $\frac{f_2}{\sum_i f_i}$, green axis to $\frac{f_1}{\sum_i f_i}$, point $N = (\frac{1}{n},...,\frac{1}{n})$ corresponds to $f_1=f_2=...=f_n$, all points to the left of $N$ - to the convex sequences $\{f_i\}$, the curved surface corresponds to the normalized left hand side, hyperplane - to the normalized right hand side.

Rigorous proof will involve a study of the geometry of the surface given by the image of the set $\{w \in \mathbb{R}^n : \sum_i w_i = 1\}$ (simplex) under the map $w \to \prod_i w_i^{w_i} + \prod_i w_i^\frac{1}{n}$.

Detailed argument: (to be finished)

  1. Slight change of notation:

\begin{equation} f_i := f(i) \quad w_i := \frac{f_i}{\sum_i f_i}. \end{equation}

Our inequality becomes

\begin{equation} \prod_i f_i ^{w_i} + \prod_i f_i^\frac{1}{n} \leq f_1 + f_n. \end{equation}

  1. Divide both sides of inequality by $\sum_i f_i$:

\begin{equation} \prod_i w_i^{w_i} + \prod_i w_i^\frac{1}{n} \leq w_1 + w_n. \end{equation}

One can recognize entropy probability ($\exp(-H(w))$ where $H(w)$ is the entropy of distribution $w$) and geometric mean in the functions on the left side.

\begin{equation} E(w) := \prod_i w_i^{w_i} \quad G(w) := \prod_i w_i^\frac{1}{n}. \end{equation}

  1. Left hand side has a critical point $\bar w = (\frac{1}{n},...,\frac{1}{n})$.
  2. Study of Hessian evaluated at this $\bar w$.
  3. Study the behavior of $L(\lambda; w^*) = E(\lambda \bar w + (1-\lambda) w^*)+G(\lambda \bar w + (1-\lambda) w^*)$ depending on $w^*$.
  4. ...some more math gymnastics...
  5. Q.E.D.!
  • $\begingroup$ Reminds me of sciencecartoonsplus.com/pages/gallery.php ... $\endgroup$ – Martin R Nov 6 '19 at 14:09
  • $\begingroup$ Haha, more or less. What do you have to say about the reasoning itself? $\endgroup$ – Konstantin Nov 6 '19 at 15:03
  • $\begingroup$ Too vague for my taste! $\endgroup$ – Martin R Nov 6 '19 at 15:08

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.