An olympiad-like inequality [closed]

Question

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 .

Answer 1

The statement at the end of the OP, that "If the function $f(x)$ is concave and positive the inequality of the beginning is reversed" seems to have a counterexample:

For $f(x)=x^{99/100}$ the inequality $$\Big(f(1)^{f(1)}f(2)^{f(2)}\cdots f(n)^{f(n)}\Big)^{\frac{1}{f(1)+f(2)+\cdots + f(n)}}+\Big(f(1)f(2)\cdots f(n)\Big)^{\frac{1}{n}}\geq f(1)+f(n)$$ is satisfied for $n=3$, but not for $n=4$ or larger. The violation is small, but within numerical accuracy:

In[1]:= f[x_] := x^(1 - 1/100)

In[2]:= N[Table[Product[f[j]^f[j],{j, 1, n}]^(1/Sum[f[j],{j, 1, n}])
          +Product[f[j],{j,1,n}]^(1/n)-f[1]-f[n],{n,3,5}],10]

Out[2]= {0.001042561976, -0.001537906443, -0.005870494687} 

"Out[2]" is the left-hand-side minus the right-hand-side of the inequality, for n=3,4,5.