Proving $\frac{1-\exp(-2st)}{s}-\frac{t}{0.5+st}\geq(\sqrt{\frac{1- \exp(-2st)}{s}} -\sqrt{ \frac{t}{1 + st}})^2$ for positive $s$ and $t$ I am trying to prove the following

$$\frac{1- \exp(-2st)}{s} - \frac{t}{0.5 + st} \geq \left(\sqrt{\frac{1- \exp(-2st)}{s}} -\sqrt{ \frac{t}{1 + st}} \right)^2 $$
for $s&gt;0$ and $t&gt;0$.

I am able to simplify it, but not able to go beyond it. Following is the simplification.
$$\frac{1- \exp(-2st)}{s} - \frac{t}{0.5 + st} \geq \frac{1- \exp(-2st)}{s} + \frac{t}{1 + st} - 2\left(\sqrt{\frac{1- \exp(-2st)}{s}}\sqrt{ \frac{t}{1 + st}} \right) \\
\Rightarrow 2\left(\sqrt{\frac{1- \exp(-2st)}{s}}\sqrt{ \frac{t}{1 + st}} \right) \geq \frac{t}{0.5 + st} + \frac{t}{1 + st} \\
\Rightarrow 4\left(\frac{1- \exp(-2st)}{st} \frac{1}{1 + st} \right) \geq \left(\frac{1}{0.5 + st} + \frac{1}{1 + st}\right)^2
 $$
The last line comes after squaring.
Okay here is my improved answer
$$\frac{1-e^{-2st}}{s} - \frac{t}{0.5 + st}\geq \left(\sqrt{\frac{1-e^{-2st}}{s}} - \sqrt{\frac{t}{1 + st}}\right)^2$$
for $s,t&gt;0$\
Simplifying the given expression we have:
$$2\left(\sqrt{\frac{1-e^{-2st}}{s}}\sqrt{\frac{t}{1 + st}}\right)\geq \frac{t}{0.5 + st} + \frac{t}{1 + st} $$
Since  $\frac{t}{0.5 + st} \geq \frac{t}{1 + st}$
\Then,
$$2\left(\sqrt{\frac{1-e^{-2st}}{s}}\sqrt{\frac{t}{1 + st}}\right)\geq\frac{t}{1 + st} + \frac{t}{1 + st}$$
$$2\left(\sqrt{\frac{1-e^{-2st}}{s}}\sqrt{\frac{t}{1 + st}}\right)\geq \frac{2t}{1 + st}$$
squaring both sides we have
$$4\left(\sqrt{\frac{1-e^{-2st}}{s}}\sqrt{\frac{t}{1 + st}}\right)^2\geq 4\left(\frac{t}{1 + st}\right)^2$$
$$\frac{1-e^{-2st}}{s}\frac{t}{1 + st}\geq \frac{t^2}{(1 + st)^2}$$
Simplifying this we have
$$1-e^{-2st} \geq \frac{st}{1+st}$$
But if $0 &lt; s &lt; t &lt; 1,$  $1- e^{-2st}&lt;0$
Which implies that the inequality is no true for $s,t&gt;0$
Okay here is my improved answer
$$\frac{1-e^{-2st}}{s} - \frac{t}{0.5 + st}\geq \left(\sqrt{\frac{1-e^{-2st}}{s}} - \sqrt{\frac{t}{1 + st}}\right)^2$$
for $s,t&gt;0$\
Simplifying the given expression we have:
$$2\left(\sqrt{\frac{1-e^{-2st}}{s}}\sqrt{\frac{t}{1 + st}}\right)\geq \frac{t}{0.5 + st} + \frac{t}{1 + st} $$
Since  $\frac{t}{0.5 + st} \geq \frac{t}{1 + st}$
\Then,
$$2\left(\sqrt{\frac{1-e^{-2st}}{s}}\sqrt{\frac{t}{1 + st}}\right)\geq\frac{t}{1 + st} + \frac{t}{1 + st}$$
$$2\left(\sqrt{\frac{1-e^{-2st}}{s}}\sqrt{\frac{t}{1 + st}}\right)\geq \frac{2t}{1 + st}$$
squaring both sides we have
$$4\left(\sqrt{\frac{1-e^{-2st}}{s}}\sqrt{\frac{t}{1 + st}}\right)^2\geq 4\left(\frac{t}{1 + st}\right)^2$$
$$\frac{1-e^{-2st}}{s}\frac{t}{1 + st}\geq \frac{t^2}{(1 + st)^2}$$
Simplifying this we have
$$1-e^{-2st} \geq \frac{st}{1+st}$$
But if $0 &lt; s &lt; t &lt; 1,$  $1- e^{-2st}&lt;0$
Which implies that the inequality is no true for $s,t&gt;0$