Question regarding one modification of one inequality of Young type

I'm reading the paper "Non - Variational elliptic systems in dimension two: a priori bounds and existence of positive solutions" by Djairo G. de Figueiredo, João Marcos do Ó and Bernhard Ruf and in Prosition 4.2 (which they use to prove Theorem 1.3) they claim that for $r,s,t,a,b \geq 1$ we have

$$rst \leq e^{t^a}+e^{r^b}+s(\ln(s))^{\frac{1}{a}+\frac{1}{b}}.$$

Unfortunately the above inequality isn't true, as can be seen by the counter-example

$$a=1, \quad b=9/8, \quad t=8, \quad r=6, \quad s=200, $$

provided by River Li in another question.

By inspecting the proof of Theorem 1.3 I've noticed that is enough the existence of some constante $C>0$ such that

$$rst \leq C(e^{t^a}+e^{r^b}+s(\ln(s))^{\frac{1}{a}+\frac{1}{b}}),$$ which is equivalent to prove that the function

$$\frac{rst}{e^{t^a}+e^{r^b}+s(\ln(s))^{\frac{1}{a}+\frac{1}{b}}}$$

is bounded for $r,s,t,a,b \geq 1$.

The problem is: Because the number of variables is $5$, I'm having problem in prove (or disprove) the above boundness, that is the reason of my question here, anyone knows how to prove (or disprove) the above boundness? any help or suggestion of techniques will be very helpful, thank you!

The answer is yes. Indeed, without loss of generality $t^a\le r^b$, so that $t\le r^{b/a}$ and hence, with $p:=\frac{ab}{a+b}=\frac1{1/a+1/b}\ge1/2$ and $x:=r^{1+b/a}\ge1$, we have $$R:=\frac{rst}{e^{t^a}+e^{r^b}+s\ln^{1/p}s} \le R_1:=\frac{r^{1+b/a}s}{e^{r^b}+s\ln^{1/p}s} =\frac{xs}{e^{x^p}+s\ln^{1/p}s}.$$ If $x\le2^{1/p}\ln^{1/p}s$, then $$R_1\le\frac{2^{1/p}s\ln^{1/p}s}{e^{x^p}+s\ln^{1/p}s}\le2^{1/p}\le4,$$ and if $x\ge2^{1/p}\ln^{1/p}s$, then $sx\le e^{x^p/2}x\le c_pe^{x^p}$ with $c_p:=(p\sqrt e/2)^{-1/p}$, so that $$R_1\le\frac{c_pe^{x^p}}{e^{x^p}+s\ln^{1/p}s}\le c_p\le16/e.$$ Thus, $$R\le R_1\le\max(4,16/e)=16/e.\quad\Box$$

Yes, this holds with some $C>0$.

With multiplicative constant, the sum of three positive numbers is the same thing as the maximum of them, which makes the things easier.

Here you may start with bounding $$e^{t^a}+e^{r^b}>(e^{t^a})^{b/(a+b)}\cdot (e^{r^b})^{a/(a+b)}=e^{t^a\cdot b/(a+b)+r^b\cdot a/(a+b)}\geqslant e^{t^{ab/(a+b)}\cdot r^{ab/(a+b)}}=e^{x^p},$$ where $x=rt\geqslant 1$, $p=ab/(a+b)=1/(1/a+1/b)\geqslant 1/2$. Here we used the usual Young (or weighted AM-GM if you prefer) inequality $\alpha X+(1-\alpha)Y\geqslant X^\alpha Y^{1-\alpha}$ for $\alpha=b/(a+b)$, $X=t^a$, $Y=r^b$.

Thus it suffices to prove $xs\leqslant C(e^{x^p}+s(\log s)^{1/p})$ for $p\geqslant 1/2$, $s,x\geqslant 1$. Consider two cases.

$s\leqslant C e^{x^p}/x$. Then $xs\leqslant Ce^{x^p}\leqslant C(e^{x^p}+s(\log s)^{1/p})$.

$s>C e^{x^p}/x$. Then $\log s>\log C+x^p-\log x$. Choose $C>2$ so large that $\log C+(1-1/\sqrt{2})\sqrt{x}-\log x>0$ for all $x\geqslant 1$, such a constant $C$ clearly exists. Then $$\log C+x^p-\log x-(x/2)^p= \log C+x^p(1-2^{-p})-\log x\\ \geqslant \log C+(1-1/\sqrt{2})\sqrt{x}-\log x>0.$$ Thus, $\log s>(x/2)^p$, and $s(\log s)^{1/p}>sx/2$, so, $$xs\leqslant 2s(\log s)^{1/p}\leqslant Cs(\log s)^{1/p}\leqslant C(e^{x^p}+s(\log s)^{1/p}).$$

Let us answer with Mathematica 14.1 on Windows

{1.27067056631641, {r -> 9.38892109031199, s -> 1617.98240176060, t -> 9.38896380431499, a -> 1.00000000000000, b -> 1.00000000000000}}

and with Maple 2024 on Windows

1.27067056647322538, [a = 1.00000000000000000, b = 1.00000000000000000, r = 9.38905609899539159, s = 1618.17799197246530, t = 9.38905609894953262]