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I want to show that

$$\left( A + \frac{t}{x}N \right)^x \le \left( A + \frac{s}{y}N \right)^y $$

or equivalently

$$x\ln\left( A + \frac{t}{x}N \right) \le y\ln\left( A + \frac{s}{y}N \right)$$

We know that $x < y$ and $t+x = T$ where $T$ is a known constant. Also $s+y < T$, all the parameters and variables are non-negative.

I tried to optimize the function $f(x) = x\ln\left( A + \frac{T-x}{x}N \right)$ to show it's maximum value is always less than right hand side, but the solution to stationary points is not solvable, in other words

$$f'(x) = \ln\left( A + \frac{T-x}{x}N \right) - \frac{\frac{T}{x}N}{A + \frac{T-x}{x}N} = 0$$ has no closed form solution for $x$. How should I proceed in comparing these two? Is my approach correct? Is there another way? (BTW if this helps, $f(x)$ is convex).

Thanks in advance

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  • $\begingroup$ Did you check it numerically? $\endgroup$
    – River Li
    Nov 13, 2021 at 10:50
  • $\begingroup$ I did plot them with MATLAB and left is less than right, but maybe my range of numbers make it true, and not in general if that's what you mean. So I should ask is this inequality true in general? $\endgroup$
    – K.K.McDonald
    Nov 13, 2021 at 10:55
  • $\begingroup$ I think it is not true in general. What's the range of $A, N, T$? $\endgroup$
    – River Li
    Nov 13, 2021 at 10:57
  • $\begingroup$ $A,N$ are random variables, $N \in [1000,5000]$ and $A \in [100,600]$ are typical range for those parameters, $T = 0.1$ is deterministic. In each snapshot, I assume we completely know $A,N$ and thus we get the deterministic inequality above. $\endgroup$
    – K.K.McDonald
    Nov 13, 2021 at 11:23
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    $\begingroup$ Please check $A = 5000, N = 600, T = 1/10, x = 9/340, y = 2/75, t = 5/68, s = 1/100$. $\endgroup$
    – River Li
    Nov 13, 2021 at 11:53

1 Answer 1

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$$f'(x) = \log\left( A + \frac{T-x}{x}N \right) - \frac{\frac{T}{x}N}{A + \frac{T-x}{x}N} = 0$$

Let $$ A + \frac{T-x}{x}N=z \implies \frac{A-N-z+z \log (z)}{z}=0$$ and the solution is given in terms of Lambert function $$z=\frac{N-A}{W\left(\frac{N-A}{e}\right)}\implies x=\frac{N\, T\, W\left(\frac{N-A}{e}\right)}{(N-A) \left(1+W\left(\frac{N-A}{e}\right)\right)}$$

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  • $\begingroup$ Thank you for heads up and answer, but I think I calculated correctly $\frac{d}{dx}\left( x\ln\left( A + \frac{T-x}{x}N \right) \right) = \ln\left( A + \frac{T-x}{x}N \right) + x\times \frac{d}{dx}\left( \ln\left( A + \frac{T-x}{x}N \right) \right) = \ln\left( A + \frac{T-x}{x}N \right) + x\times \frac{\frac{-T}{x^2}N}{A + \frac{T-x}{x}N}$ BTW, If the answer can be determined in terms of Lambert-W, how can I compare LHS with RHS with the solution? Thanks a lot for Hint. $\endgroup$
    – K.K.McDonald
    Nov 13, 2021 at 10:11
  • $\begingroup$ @K.K.McDonald. You are correct but my result stays the same. Cheers and sorry ! $\endgroup$
    – Claude Leibovici
    Nov 13, 2021 at 10:30
  • $\begingroup$ Claude Leibovici, thank you for the answer, can you please explain how I use it to compare RHS with LHS? My main question was with regard to inequality I'm am trying to solve. $\endgroup$
    – K.K.McDonald
    Nov 13, 2021 at 10:44

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