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.
 A: 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 $x>0$.
A: You can show fairly simply that the first term will be negative and the second term positive with positive x and 1 > t > 0. 
Because they are added, showing that the overall function is negative becomes a question of showing that the first (negative) term dominates, which you can do by setting up and solving an inequality between them and show that it holds given the restrictions on x and t.
