Lower Bound for $\frac{t + (1-t)p_1}{1+2t}\cdot \frac{t + (1-t)p_2}{1+t}$ I am interested in obtaining a lower bound for the function
$$
f(t) = \frac{t + (1-t)p_1}{1+2t}\cdot \frac{t + (1-t)p_2}{1+t},
$$
defined on $(0,1)$, where $p_1$ and $p_2$ are two constants with the constraint that $p_1 + p_2 \leq 1$ and they are both strictly positive. Playing around with a graphing tool I have convinced myself that a lower bound is given by $\frac{2}{3}p_1p_2$, and that this bound is sharp. (Here, I mean that for any choice of $p_1, p_2$ the value is a lower bound, even though it might not necessarily be achieved for arbitrary $p_1$ and $p_2$. But taking $p_1 = p_2 = \frac{1}{2}$ and letting $t \to 1$ we get a function value of $\frac{1}{6} = \frac{2}{3}p_1p_2$, so in general we for sure can't find a better bound without more assumptions.)
Unfortunately, I struggle with formally showing that this gives in fact a lower bound.
I tried assuming that $p_2 \leq \frac{1}{2}$ (since by the condition not both parameters can be larger than $\frac{1}{2}$ and then using the fact that $1-p_2 \geq p_2$. With that, I can bound the factor on the right by $p_2$, but I suspect that this method is too loose since then I have no way to bound the left factor.
Maybe someone sees a fairly straight-forward way to show the lower bound? Thanks!

UPDATE
Since I am really trying to avoid going down the Calculus route, i followed @Thomas Andrews suggestions from the comments and looked at $f(t) - \frac{2}{3}p_1p_2$ in the hopes of being able to show that the resulting numerator is always positive. I have made the following progress.
We can compute
$$
f(t) - \frac{2}{3}p_1p_2 = -\frac{3p_1t^2 + p_1p_2t^2 + 3p_2t^2 - 3p_1t + 12p_1p_2t - 3p_2t - p_1p_2 - 3t^2}{3(t+1)(2t+1)}.
$$
Therefore we can look at the numerator:
$$
3t^2 - 12p_1p_2t+3p_1t-3p_1t^2 + 3p_2t - 3p_2t^2 +\underbrace{p_1p_2 - p_1p_2t^2}_{\geq 0}.
$$
Since the last two summands together are nonnegative, we can proceed to show that the first six summands together are nonnegative. These first six terms factor as
$$
3(1-(p_1+p_2))t^2 + 3t(p_1+p_2 - 4p_1p_2),
$$
and since $p_1+p_2 \leq 1$ the first summand is non-negative. We have thus reduced the problem to showing that
$$
p_1+p_2-4p_1p_2 \geq 0,
$$
where the only thing we really have at our disposition is the fact that $p+q \leq 1$. I suspect this to be true, but seem to struggle with a proof. But this seems like it should not be a big deal. Anyone seeing how to show this?

UPDATE 2 Thanks to @Svyatoslav, here is a way to finish the argument.
Let us write $p = s^2, q = t^2$, so that we have $s^2 + t^2 \leq 1$. We can parametrize $s$ and $t$ by $s = r\cos(\phi), t = r\sin(\phi)$, so that we have $s^2 + t^2 = r^2 \leq 1$. Using this and the double angle formula, we can then obtain
$$
p+q-4pq = r^2 - 4r^4\sin^2\phi\cos^2\phi = r^2 - r^4\sin^2 (2\phi) \geq r^2 - r^2\sin^2 (2\phi) = r^2\cos^2(2\phi) \geq 0.
$$
This shows that indeed, $\frac{2}{3}p_1p_2$ is a lower bound of $f$.
 A: Assume that $p_1, p_2 \ge 0$.
We have
\begin{align*}
 f(t) &= \frac{(1 - p_1)(1 - p_2)}{2} - \frac{1 - p_1 - p_2}{2}\cdot \frac{1 + 3t}{(1+t)(1+2t)}\\[6pt]
 &\qquad + \frac{p_1p_2}{2} \cdot \frac{1}{1+2t}
 + (p_1 + p_2 - 4p_1p_2)\cdot \frac{t}{(1+t)(1+2t)}\\[8pt]
 &\ge \frac{(1 - p_1)(1 - p_2)}{2} - \frac{1 - p_1 - p_2}{2} \cdot 1 + \frac{p_1p_2}{2} \cdot \frac{1}{3}\\[6pt]
 &= \frac23 p_1p_2
\end{align*}
where we have used $\frac{1 + 3t}{(1+t)(1+2t)} = \frac{1+3t}{1+3t+2t^2}\le 1$ and $p_1 + p_2 - 4p_1p_2
\ge p_1 + p_2 - (p_1 + p_2)^2 \ge 0$.
A: If the $p_i$ are allowed to be negative, this argument doesn't work.
If $p_i$ are positive:
$$f(t)=p_1p_2\left(\frac{1+\frac{1-p_1}{p_1}t}{1+2t}\cdot \frac{1+\frac{1-p_2}{p_1}t}{1+t}\right)$$
Letting $$q_i=\frac{1-p_i}{p_i}=\frac{1}{p_i}-1\geq 0,$$ then $p_i=\frac{1}{q_i+1}$ and $p_1+p_2\leq 1$ is equivalent to $1\leq q_1q_2.$
Then this becomes $$f(t)=p_1p_2\frac{(1+q_1t)(1+q_2t)}{(1+t)(1+2t)}$$
So you want to show, for $q_1,q_2>0,$ and $q_1q_2\geq 1,$ and $t\in[0,1],$ that:
$$3(1+q_1t)(1+q_2t)\geq 2(1+t)(1+2t)$$
This can be written as:
$$1+3(q_1+q_2-2)t+(3q_1q_2-4)t^2\geq 0\tag1$$
But $q_1q_2\geq 1$ means $q_1+q_2\geq 2,$ by AM-GM, so the only way the left side of $(1)$ can be negative is  is if $q_1q_2< \frac43.$
Then the left side of $(1)$ has no local minimum, so the only place where it can be minimized is at $t=0$ or $t=1.$
When $t=0,$ the values is $1,$ so there it is positive.
When $t=1,$ the value is:
$$1+3(q_1+q_2-2)+(3q_1q_2-4)=3(q_1q_2+q_1+q_2)-9=3(q_1+1)(q_2+1)-12$$
So to find a contradiction, you need to have $1<q_1q_2<\frac43$ and $$(1+q_1)(1+q_2)<4.$$
But $1+q_1+q_2+q_1q_2>2+q_1+q_2\geq 2+2\sqrt{q_1q_2}>4.$
So we get $f(t)\geq \frac23p_1p_2.$

If the assumptions is the $p_i$ are non-negative, then we have to handle $p_1=0$ or $p_2=0.$

If the denominator is $(1+at)(1+bt)$ for $a,b>0$ and you want a minimum of $\alpha p_1p_2,$ then you want:
$$(1+q_1t)(1+q_2t)\geq \alpha(1+at)(1+bt)$$
or $$(1-\alpha)+(q_1+q_2-\alpha(a+b))t+(q_1q_2-\alpha ab)t^2\geq 0.\tag2$$
So you need $\alpha<1,$ or $(2)$ is negative when $t=0.$
Certainly, if $\alpha(a+b)\leq 2,$ then $q_1+q_2-\alpha(a+b)>0$$
Then if $\alpha ab \leq q_1q_2,$ then we always get non-negative coefficients, so $$\alpha =\min\left(\frac2{a+b},1,\frac{1}{ab}\right)$$
we certainly get our needed inequality.
But as before, we might allow the quad ration coefficient to be negative. Again, it has not local minimum in that case, so the minimum is either $1-\alpha$ or $$(1+q_1)(1+q_2)-\alpha(1+a)(1+b).$$ For the latter to be positive, we need $$\alpha\leq\frac{(1+q_1)(1+q_2)}{(1+a)(1+b)}.$$
It is enough to have $$\alpha= \min\left(1,\frac4{(1+a)(1+b)}, \frac2{a+b}\right).$$
This gives $\frac23$ when $a=2,b=1.$
When $b=1, a\geq 1$ the two terms are equal, and we get: $\alpha=\frac2{a+1}$ works.
