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May 10

Solving Inequality with Polygarithms

I am looking for a (preferably simple) function $f$, and a constant $r\in (0,1)$ that satisfy $f(x,y,k)=\mathrm{\Omega }\left(\right(\frac{\log (y)}{x^r}{)}^k)$, and also $\underset{a,b,c}{\min }\{f(a,b,k)+f(d-a,c,k)+f(d,e-bc,k)\}=\mathrm{\Omega }(f(d,e,k))$.

So, I came up with $f(x,y,k)=(\frac{\log (y)}{{\log }^r(x)}{)}^k$ that I think that it satisfies even stornger condition: $f(a,b,k)+f(d-a,c,k)+f(d,e-bc,k)\ge f(d,e,k)$ for all $a,b,c,d,e,k$ in $f$'s definition domain (but I can't neither prove it, nor finding the desired $r>0$).

Does it work for my function?

Do you have any simpler function, with simple proof that it works?

Any help would be highly appreciated! David (talk) 13:55, 10 May 2019 (UTC)

Here ${\log }^r(x)$ means $(\log (x){)}^r$? One minor observation is that the function is concave up in its first coordinate, so (assuming that you expect all arguments always to be positive, or perhaps even larger than 1) the minimum of the LHS is always obtained when $a=d/2$. --JBL (talk) 19:09, 13 May 2019 (UTC)