Can a function be bounded from below by its second order Taylor expansion? Define the function $f:[0,1) \rightarrow [0,\infty) $ given by 
$$ \large{f(\theta)} := (1-\theta^2)^p (1+\theta)^{p \theta} (1-\theta)^{-p \theta}.$$ 
If $p$ is a sufficiently large positive real number, is it true that 
$$ \large{f(\theta) \geq C(1+p \theta^2)}$$ 
for some positive constant $C$? 
To clarify, the constant $C$ is independent of $p$ and $\theta$? 
Note that upto second order $f(\theta)$ is equal to $(1+p\theta^2)$. To see why, 
note that 
$$ (1-\theta ^2)^p = 1- p \theta ^2 + O(\theta^3), \qquad 
(1+\theta)^{p \theta} = 1+ p \theta^2 + O(\theta^3), \qquad  
(1-\theta)^{-p \theta} = 1+ p \theta^2 + O(\theta^3).  $$ 
This proves that upto second order $f(\theta)$ is equal to $(1+p\theta^2)$. 
Remark: If one can show 
$$(1+\theta)^{1+\theta} (1-\theta)^{1-\theta} \geq 1+\theta^2,$$ 
then we are done, via Bernouli's inequality. 
This was suggested by someone, in a now deleted post. This inequality does seem to be true (I tried plugging in $\theta = 0.1, 0.2, \ldots, 0.9$....it holds true). 
Note that this inequality doesn't depend on $p$, hence its probably more tractable. 
 A: For your last inequality(in your remark), let $x\in (0,1)$, you have::
$$(1+x)\log(1+x)=x+\sum_{n\geq 2}\frac{(-1)^n}{n(n-1)}x^n$$
Hence
$$(1+x)\log(1+x)+(1-x)\log(1-x)=\sum_{n\geq 2, 2|n}\frac{2x^n}{n(n-1)} \geq x^2$$
Thus:
$$(1+x)^{1+x}(1-x)^{1-x}\geq \exp(x^2)\geq 1+x^2$$ 
A: Since
$$\frac{d^2}{dx^2} \ln\left(\frac{1+x}{1-x}\right) = \frac{4x}{(1-x^2)^2} $$
we have that $\ln\bigl(\frac{1+x}{1-x}\bigr)$ is convex on $[0,1)$.  So, if $x\in[0,1)$ and $\lambda\in[0,1]$, then
$$ \ln\left(\frac{1+((1-\lambda)0 + \lambda x)}
  {1-((1-\lambda)0 + \lambda x)}\right)
\le (1-\lambda)\ln\left(\frac{1+0}{1-0}\right)
+\lambda \ln\left(\frac{1+x}{1-x}\right) 
$$
Simplifying, we get
$$ \frac{1+\lambda x}{1-\lambda x}
\le \left(\frac{1+x}{1-x}\right)^\lambda
\qquad\text{if $x\in[0,1)$ and $\lambda\in[0,1]$.}
\tag1
$$
(This is reminiscent of Bernoulli's inequality, but doesn't seem to imply it or be implied by it.)  In particular, taking $\lambda=x\in[0,1)$, we get
$$ \frac{1+x^2}{1-x^2}
\le \left(\frac{1+x}{1-x}\right)^x
\qquad\text{if $x\in[0,1)$.}
$$
Multiplying by $1-x^2$ yields the desired inequality
$$ 1+x^2 \le (1+x)^{1+x}(1-x)^{1-x} $$

Remark: the convexity of $g(x)=\ln\bigl(\frac{1+x}{1-x}\bigr)$ can also be established without calculus.  Indeed, if $a>b$ then $t\mapsto\frac{a+t}{b+t}$ is a decreasing function of $t$; since $\bigl(\frac{x+y}2\bigr)^2\ge xy$ by AM/GM, this yields
$$ \frac{1+x+y+\tfrac14(x+y)^2}{1-x-y+\tfrac14(x+y)^2}
\le \frac{1+x+y+xy}{1-x-y+xy} $$
That is,
$$ \frac{\bigl(1+\tfrac12(x+y)\bigr)^2}{\bigl(1-\tfrac12(x+y)\bigr)^2}
\le \frac{(1+x)(1+y)}{(1-x)(1-y)} $$
Taking square roots, then logs, yields $g(\frac12(x+y))\le\frac12(g(x)+g(y))$, which since $g$ is continuous establishes that $g$ is convex.

An additional remark: If $a$ and $b$ are positive then
$$ \frac{a+b}2 \le \sqrt{\frac{a^2+b^2}{2}} \le a^{a/(a+b)} b^{b/(a+b)} $$
The first inequality is the standard QM/AM inequality.  The fact that the far left is less than the far right is an instance of the standard GM/HM inequality.  The second inequality is a variant of yours: take $x=\frac{a-b}{a+b}$.  In this form, your inequality fits very nicely into the general scheme of standard mean inequalities.
