Question on function I have to prove this affirmation :

the function $a : [0; 1] \rightarrow[0; 1]$ defined by $a(t) =\displaystyle\frac{t(2p-t)}{p^2}$
satisfied that
$a(t)\geq \min\lbrace t; 1 -t\rbrace $ for $t\in[0; 1]$.
such that $\frac12<p<1$

So i have to prove that :$a(t) \geq t$ ,if $t<1-t$ and $a(t) \geq 1-t$ ,if $t>1-t$
but i can't prove that: 
$a(t) \geq t$ ,if $t<1-t$ ,
for the second : 

$\frac{t(2p-t)}{p^2} \geq 1-t \Rightarrow t(2p-t)\geq p^2-tp^2
 \Rightarrow tp-t^2-p^2+tp^2\geq 0 \Rightarrow (p-t)^2+tp^2\geq 0$ this
  is true for all $t$.

so i just need help for the first !
Please help me 
Thank you .
 A: Ok, with the edit I understand now.  So, for the first part then we're assuming that $\min(t,1-t) = t$.  Since we want to show that $a(t) \geq \min(t,1-t)$ this is now the same thing as showing that $a(t)-t \geq0$.  So:
$$\frac{t(2p-t)}{p^2} -t \geq 0$$
$$\Leftrightarrow 2tp - t^2 -tp^2 \geq 0 \mbox{ since $p^2>0$}$$
$$\Leftrightarrow -t + (2p-p^2) \geq 0 \mbox{ assuming $t\not=0$}$$
$$\Leftrightarrow 2p-p^2 \geq t \ \ \forall t \in \left(0,\frac{1}{2}\right)$$
Note that we exclude $0$ from the interval for $t$ now since we explicitly excluded the case $t=0$ above; we'll return to that at the end.  Note also that the interval only goes to $\frac{1}{2}$ because $t<1-t$. We know that $\frac{1}{2}<p<1$, so can we now conclude that $2p-p^2 \geq t$?
Well, $2p-p^2$ is increasing on $\left(\frac{1}{2},1\right)$ so we only need look at the value of the left endpoint.  When $p=\frac{1}{2}$ we have $2p-p^2 = \frac{3}{4}$ which is greater than $\frac{1}{2}$ which is as large as $t$ can get.
So it is true that $2p-p^2 \geq t \ \ \forall t \in \left(0,\frac{1}{2}\right)$ and so $a(t)-t \geq 0$ as required.
Now, the edge cases:  when $t=0$ we have $a(0)-0 = 0$ so our inequality still holds for this case, so we're ok there.
For $t=\frac{1}{2}$ our proof also still holds because $2p-p^2 \geq \frac{3}{4}$, so when $\min(t,1-t)=t=1-t$ we're also still good.
