A two-point inequality Let $M(p,q) = (2p-\sqrt{p^{2}+q^{2}})\sqrt{p+\sqrt{p^{2}+q^{2}}}$ and set $B(t) = M(x+t, \sqrt{t^{2}+(y+bt)^{2}})$. Given any real $x,y,b$ is it true that $\varphi(t) = B(t)+B(-t)$ is decreasing  in $t$ for  $t \geq 0$. 
Motivation:
Consider the hamming cube $\{-1,1\}^{N}$. Let $f :\{-1,1\}^{N} \to \mathbb{R}$. Set $f_{k} = \mathbb{E} (f| \mathcal{F}_{k})$ to be a martingale $k=0,..,N$  which takes the average of the function with respect to the variables $(x_{k+1},...,x_{N})$. So $f_{0} = \mathbb{E} f = \frac{1}{2^{N}} \sum_{x \in \{-1,1\}^{N}} f(x)$, $f_{N} := f$. So $f_{k}$ lives on $\{-1,1\}^{k}$. For example 
$$
f_{N-1}(x) = \frac{1}{2}\left(f(x_{1},\ldots, x_{N-1},1)+f(x_{1},\ldots, x_{N-1},-1) \right)
$$
Define $\nabla_{i} f := \frac{1}{2}\left(f(x_{1},x_{2},\ldots, 1,\ldots, x_{N}) -f(x_{1},x_{2},\ldots, -1,\ldots, x_{N})\right)$. And let $|\nabla f|^{2} = \sum_{i=1}^{N}|\nabla_{i} f|^{2}$. Now let $T_{\rho}$ be the Ornstein-Uhlenbeck semigroup on $\{-1,1\}^{N}$ i.e., 
$$
T_{\rho} f = \sum_{S \in 2^{N}} \rho^{|S|} \hat{f}(S) W_{S}(x) 
$$
Where $W_{S}(x)$ is the Walsh system , and $\hat{f}(S)$ are Fourier coefficients with respect to this system. My question becomes the claim that the following map 
$$
\rho \to \mathbb{E}M(T_{\rho}f_{k}, |\nabla T_{\rho} f_{k}|)
$$
is monotone for $\rho \in [0,1]$ and any $k \geq 1$. This in particular makes the process $M(f_{k}, |\nabla f_{k}|)$  supermartingale. And then central limit theorem $N \to \infty$ gives some interesting inequalities. 
 A: The following may be the starting point of an answer. 
Introduce the new variables
\begin{equation}
 u:=\sqrt{(y-b t)^2+t^2+(t-x)^2},\quad v:=\sqrt{(b t+y)^2+t^2+(t+x)^2}, 
\end{equation}
so that $b=\frac{-4 t x-u^2+v^2}{4 t y}$. Further let 
\begin{equation}
 r:=\sqrt{-t+u+x},\quad s:=\sqrt{t+v+x}, 
\end{equation}
so that 
\begin{equation}
 u = r^2 + t - x, \quad v = s^2 - t - x. 
\end{equation}
Then 
\begin{equation}
 \varphi'(t) \frac{32}3\, r s t y^2
 =-\left(r^2+t-x\right)^2 \left(-2 (r+s) \left(-s^2+t+x\right)^2+8 t x (r+s)+4 y^2
   (s-r)\right)
\end{equation}
\begin{equation}   
   -16 s t y^2 \left(r^2+t-x\right)-(r+s) \left(r^2+t-x\right)^4-16
   r t y^2 \left(-s^2+t+x\right)
\end{equation}
\begin{equation}   
   +4 \left(-s^2+t+x\right)^2 \left(2 t x (r+s)+y^2
   (s-r)\right)-(r+s) \left(-s^2+t+x\right)^4
\end{equation}
\begin{equation}   
   -16 t x \left(t x (r+s)+2 y^2
   (s-r)\right),  
\end{equation}
which is a polynomial in $x,y,r,s,t$. 
We need to show that this polynomial is nonnegative on an appropriate set (say $S$) of 5-tuples $(x,y,r,s,t)$. 
In principle, this can be done by using standard methods of real algebraic geometry. However, I have not yet succeeded in obtaining a tractable description of such a set $S$. 
