<s>[INST] Show that a function is negative over its domain I would like to demonstrate that the following function is negative
\begin{equation}
f(x)=-\frac{t}{4\sqrt{x}^3}\bigg[1-\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}}\bigg]+\bigg(\frac{1}{1-t}\bigg)\bigg(\frac{t}{2\sqrt{x}}\bigg)^2\bigg(1+\sqrt{x}\bigg)^{\frac{2-t}{t-1}}
\end{equation}
for $x>0$ and $1>t>0$.
I have graphed this in Desmos and confirmed that the function is negative given my restrictions on $x$ and $t$.
Any help to solve this analytically would be most appreciated!
Many thanks.
 [/INST] Here is an attempt to answer my question:
The function $f(x)$ has the same sign as the function $g(x)$, where
\begin{equation*}
    g(x)=\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}}\bigg[1+\frac{t}{1-t}\frac{\sqrt{x}}{(1+\sqrt{x})}\bigg]-1,
\end{equation*}
and so the problem is equivalent to showing that $g(x)<0$.
Note that $\lim_{x\rightarrow0}g(x)=0$.
Moreover, $g(x)$ is strictly decreasing.
The first-order condition is
\begin{multline*}
    g'(x)=\frac{1}{t-1}\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}-1}\bigg(\frac{1}{2\sqrt{x}}\bigg)\\
  +\frac{t}{1-t} \Bigg[\frac{\frac{1}{2\sqrt{x}}(1+\sqrt{x})-\sqrt{x}\big(\frac{1}{2\sqrt{x}}\big)}{(1+\sqrt{x})^2}\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}}\\
      +\frac{1}{t-1}\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}-1}\bigg(\frac{1}{2\sqrt{x}}\bigg)\frac{\sqrt{x}}{(1+\sqrt{x})}\Bigg],
  \end{multline*}
\begin{multline*}
    =\frac{1}{2\sqrt{x}(t-1)}\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}-1}\\
  +\frac{t}{1-t}\Bigg[\frac{1}{2\sqrt{x}(1+\sqrt{x})^2}\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}}\\
     +\frac{\sqrt{x}}{2\sqrt{x}(t-1)(1+\sqrt{x})}\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}-1}\Bigg].
  \end{multline*}
Simplifying further yields
\begin{equation*}
    g'(x)=-\frac{1}{2\sqrt{x}(1-t)}\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}-1}+\bigg[\frac{t}{2\sqrt{x}(1-t)}-\frac{t\sqrt{x}}{2\sqrt{x}(1-t)^2}\bigg]\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}-2}.
\end{equation*}
Then, using 
\begin{equation*}
    \frac{1}{2\sqrt{x}(1-t)}=\frac{(1-t)(1+\sqrt{x})}{2\sqrt{x}(1-t)^2(1+\sqrt{x})} \ \text{ and } \ \frac{t}{2\sqrt{x}(1-t)}=\frac{t(1-t)}{2\sqrt{x}(1-t)^2},
\end{equation*}
implies that 
\begin{align*}
  g'(x)&=-\Bigg[\frac{(1-t)(1+\sqrt{x})}{2\sqrt{x}(1-t)^2(1+\sqrt{x})}\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}-1} -\bigg[\frac{t(1-t)}{2\sqrt{x}(1-t)^2}-\frac{t\sqrt{x}}{2\sqrt{x}(1-t)^2}\bigg]\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}-2}\Bigg],\\
  &=-\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}-2}\bigg[\frac{1-t+\sqrt{x}-t\sqrt{x}-t+t^2+t\sqrt{x}}{2\sqrt{x}(1-t)^2}\bigg],\\
  &=-\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}-2}\bigg[\frac{1-2t+\sqrt{x}+t^2}{2\sqrt{x}(1-t)^2}\bigg],\\
  &=-\frac{(1+\sqrt{x})^{\frac{1}{t-1}}}{(1+\sqrt{x})^2}\bigg[\frac{(t-1)^2+\sqrt{x}}{2\sqrt{x}(1-t)^2}\bigg]<0.
\end{align*}
Therefore, $g(x)$ is a strictly decreasing function starting from a limit value of $0$,
which implies the original function is negative for $x>0$.
</s>