Suppose that $(x,y,z)\in[0,1]^3$ for simplicity. It is easy for me to interpret "$x+y<z$": this just means "$z$ is larger than the sum of x and y." Obviously "$e^{x+y}<e^z$" has the same interpretation. I'm having trouble interpreting "$e^x+e^y<e^z.$" Any suggestions for an intuitive interpretation of this inequality would be greatly appreciated!

An observation: Plotting $\left\{(x,y,z)\in[0,1]^3:e^x+e^y<e^z\right\}$ in MATLAB seems to suggest that in order to be in this set, $z$ must be larger than $x+y$ plus some additional buffer. Intuition behind why this happens?

  • $\begingroup$ that observation is not actually true. "z must be larger than x+y" is the correct observation for $e^{x+y} < e^z$ and therefore the "buffer" is precisely the difference between $e^{x+y}$ and $e^x+e^y$. Setting $x=y=c$ and checking the extremes $c=0,1$, we note that $e^2 >2e^1$ but $e^0 < 2e^0$, so the "buffer" flips sign somewhere. $\endgroup$ Oct 31 '20 at 2:26
  • $\begingroup$ It seems to be true for $x,y,z$ in $[0,1]$ but fails for larger $z$. Consider just the equation $2e^x = e^{2x}$ which has solution $e^x=2$, $z=\ln 4$ slightly more than $1$. $\endgroup$ Oct 31 '20 at 2:30
  • $\begingroup$ $e^{x}+e^y <e^z$ is certainly not true when $\max(x,y) \ge z$ but it is true when $\max(x,y) +\log_e(2) \lt z$ $\endgroup$
    – Henry
    Oct 31 '20 at 2:40
  • $\begingroup$ I'm not understanding your issue. $e^x$, $e^y$ and $e^z$ are a values and if $e^x + e^y < e^z$ then ... it is so. What's to explain? If $e^x = M \in [1, e]$ and $e^y=N \in [1,e]$ and $e^z= K \in [1,e]$ and $M +N < K$ ... what needs explaining? $\endgroup$
    – fleablood
    Oct 31 '20 at 2:53
  • $\begingroup$ "that observation is not actually true." Well, neither is $x + y < z$ (what if $x=7.9; y = 8.6$ and $z =0.04$?) nor is $e^{x+y} < e^z$ (ditto). $\endgroup$
    – fleablood
    Oct 31 '20 at 2:55

the whole thing with equality is a surface. Here is a picture of level curves of $e^x + e^y$

Note that the level curves are identical, take the one that goes through the origin and shift by vector $(t, t).$ In the picture use $(\frac{1}{2}, \frac{1}{2}).$

enter image description here

Here it is with some diagonal lines drawn. If you tilt your head $45^\circ$ so that the diagonal lines appear to be horizontal, the idea that the curves are identical becomes more believable. I do not know how to rotate a jpeg or png by $45^\circ$

enter image description here

enter image description here

  • $\begingroup$ Here is it --- however I didn't see how I could insert the image in a comment to be immediately visible [![image][1]][1] [1]: i.stack.imgur.com/SB8fT.png $\endgroup$ Nov 8 '20 at 9:16
  • $\begingroup$ @GottfriedHelms got it, there is a choice from the image icon to paste from a link, that worked very well $\endgroup$
    – Will Jagy
    Nov 8 '20 at 15:59

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