Let $a_k$ and $b_k$ be ascending positive numbers for $1\leq k \leq K+1$. If it is known that $$\frac{K\left(\exp\left(\frac{1}{K}\sum_{k=1}^K b_k\right)-1\right)}{\left(\sum_{k=1}^K \sqrt{a_k} \sqrt{\exp(b_k)-1} \right)^2}=1$$ I need to prove that $$\frac{(K+1)\left(\exp\left(\frac{1}{K+1}\sum_{k=1}^{K+1} b_k\right)-1\right)}{\left(\sum_{k=1}^{K+1} \sqrt{a_k} \sqrt{\exp(b_k)-1} \right)^2}<1.$$

From Matlab this surely holds, but I do not know why.

$\mathbf{\rm Alternative \;\,formulation.\;}$ Here is an alternative formulation of the same thing: With ascending $a_k$ and $b_k$, define $$z_m=\frac{\prod_{k=1}^{m}(1+b_k)^{(1/m)}-1}{\frac{1}{m}\left[\sum_{k=1}^m\sqrt{a_kb_k}\right]^2}.$$

Let $m_0$ be the smallest $m$ such that $z_m<1$. Then, for $m>m_0$, the sequence $z_m$ is monotonically decaying. It is easy to find counter examples where the sequence increase for $z_m$ with $m<m_0$.

$\mathbf{\rm 2^{nd}Alternative \;\,formulation, \;\,continuous\;\,version.\;}$ For positive monotonically increasing functions $A(x)$ and $B(x)$, define the function $$z(T) =\log\left(1+\frac{1}{T}\left[\int_0^{T} \sqrt{A(x)B(x)}\mathrm{d}x\right]^2\right)-\frac{1}{T}\int_0^T\log(1+B(x))\mathrm{d}x.$$

Let $T_0$ be the smallest $T$ such that $z(T)>0$. Then, $z(T)>0$ for $T>T_0$.


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    $\begingroup$ It would perhaps make sense to modify the question into: Does it exist a sequence $a_k$ so that the 1st formulation holds $\endgroup$ Nov 6, 2016 at 15:37

1 Answer 1


The assertion is wrong, so this problem should be closed. The following is a counter example: $$K=2$$ $$a=[0.0005\; 0.0118\; 0.0341]$$ $$b=[14.9955 \; 25.2616\; 25.3731]$$

This disproves the first formulation of the problem. If the 2 others are equivalent, they are also wrong.


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