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 .

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 !

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

Explanation:

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.

• Will the op move the goal post by excluding $n=3$ again? – WE Tutorial School Nov 3 '19 at 17:41
• Or add some constraints on $f(1), f(2), \cdots, f(n)$. – River Li Nov 4 '19 at 1:49
• @WETutorialSchool: And indeed – the question has changed again :) – Martin R Nov 5 '19 at 10:32
• @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. – River Li Nov 6 '19 at 13:17
• @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 $\alpha > 0$. – River Li Nov 6 '19 at 16:19

Hint: Normalize both sides by $$\sum_i f(i)$$ and take a look at case $$n=3$$: 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.!
• Reminds me of sciencecartoonsplus.com/pages/gallery.php ... – Martin R Nov 6 '19 at 14:09
• Haha, more or less. What do you have to say about the reasoning itself? – Konstantin Nov 6 '19 at 15:03
• Too vague for my taste! – Martin R Nov 6 '19 at 15:08