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 .

• You said you found this inequality in a paper. Was it proved there? Can you share the paper and/or the proof? – Iosif Pinelis Oct 24 '19 at 13:54
• I think maybe "paper" means competition exam paper? If so, this is not suitable for this site. – Anthony Quas Oct 24 '19 at 23:10
• MSE is a right place for such type questions. – user64494 Oct 25 '19 at 5:36
• @user64494 It seems that the question is already posted there: An olympiad-like inequality. – Martin Sleziak Oct 25 '19 at 6:07
• As pointed out here, this seems to be wrong for $f(x) = x$ and $n=2$. – Martin R Oct 25 '19 at 6:34

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.