Inequality for $s\in(1,2)$, $t>1$

Question

Let $s\in(1,2)$, $t>1$ and $t-1>\frac{4}{q}t(s-1)$ where $q\in(0,p-1)$ for $p\geq 2$. Then does the following inequality hold: $$ (s-1)<c(p)t^q-d(p)s^p $$ for some constants $c(p),d(p)$ depending on $p$. In fact it hods when $p=2$. The proof goes as follows: By the mean value theorem there exists $\eta\in(1,t)$ such that $t^q-1=q\eta^{q-1}(t-1)$. Then $$ \frac{(s+2)(s-1)}{t-1}\leq\frac{q(s+2)}{4t}\leq\frac{q}{t}\leq\frac{q}{t^{1-q}}\leq q\eta^{q-1}=\frac{t^q-1}{t-1}.$$ Therefore, we have $$ (s+2)(s-1)\leq t^q-1, $$ which gives $$ s-1\leq t^q-s^2. $$ Hence the proof follows. So $c(p)=d(p)=1$ here. Any hint when $p>2$? Thanks in advance. $$

Answer 1

Suppose $s \in (1,2)$ , $t>1$ & $t-1 > \frac{4t(s-1)}{q}$, where $q \in (0, p-1)$ for all $p> 2$. We proceed by way of contradiction. Assume the inequality doesn't hold for any constants $c(p)$ & $d(p)$.

Let $c(p) = d(p) = p$ so that $s-1 \geq p(t^q - s^p)$. Since $s>1 $ , $p> 2$ & $t>1$ we've,

$2-1>s-1 \geq p(t^q - s^p)> 2(1- s^p)$. So, $1> 2(1- s^p)$. This gives, $s^p>1/2$ for all $p>2$ & $s \in (1,2)$. But, this is a contradiction since we can choose $s$ & $p$ such that $s^p < 1/2$.

Thus, the inequality holds for $p>2$.