Now I know the normal method of manipulation which will get us $$c+1\le 2(e^x + \frac{1}{e^x})$$ ie. $c\le 3$
But can I do it by assume $e^x=t$ and then resolving the quadratic? What complications would $t>0$ bring into it? I realise that setting $D<0$ will get the same answer but that doesn’t always work so I want to know how to adjust for $t>0$ or any other condition on the variable in a general function
I am straring with $$c+1\le 2(e^x + \frac{1}{e^x})$$ which you wrote in your attempt, but there is an extra $p$ factor in the title.
Hints: The inequality we need is $t^{2}+1-(c+ 1) t/2 \geq 0$ for all $t >0$. The left side is minimized when $t=\frac {c+1} 4$. You have to consider the cases $ c+1>0$ and $ c+1\leq 0$ separately.In the second case the inequality is automatically satisfied so there is no extra condition on $c$. In the first case the condition is $t^{2}+1-(c+ 1) t/2 \geq 0$ for $t=\frac {c+1} 4$. Can you now show that the answer is $(-\infty, 3]$?
