I want to show that 
$9\*\left[\frac{xy}{x+y}-q(1-q)\right]-12\*[xy-q(1-q)]+(1-q-x)^{3}+(x+y)^{3}+(q-y)^{3}-1\geq0$ where 


$0<q<1$


$0<x<1-q$


$0<y<q$


$(x+y)\left[1+max\{\frac{1-q}{y},\frac{q}{x}\}\right]\leq3$


I play with it numerically. It is right. But don't know how to prove it analytically. Anybody can help? Thanks a lot.



Want to prove an inequality
[Iosif's answer](https://mathoverflow.net/a/330591) is very interesting and if anyone knows how to do the same thing in Maple I'd like to know.


However I disagree with Iosif about the difficulty. Mathematica will use a systematic procedure that is guaranteed to work in a wide variety of cases, and that may take many more steps than an *ad hoc* method devised by a human.


So here is a quick proof. Remove the upper bounds on $q$, $x$, $y$ by substituting 
$x=(1-q)X/(1+X)$, $y=qY/(1+Y)$, $q=Q/(1+Q)$. After clearing demoninators that are obviously positive, we have to prove $\Phi\ge 0$ subject to $Q,X,Y\ge 0$ and
conditions $C\_1$ and $C\_2$, where
$$\begin{align}\Phi &= -Y^2(1+X)^3 Q^4 + XY(1+X)^2(Y+2)Q^3\\
 &{\quad}+ 2XY(1+Y)(1+X)(X+Y+3)Q^2 
 + XY(1+Y)^2(X+2)Q - X^2(1+Y)^3\end{align}$$
and $C\_1$, $C\_2$ come from the OP's last condition. They can be arranged like this:
\begin{align\*}
C\_1:\qquad & YQ \ge
 \frac{(1+Y)X}{(X+2)}-\frac{Y(2Y+3)(1+X)}{(1+Y)(X+2)}Q^2.
\\
C\_2:\qquad & YQ^2 \le \frac{X(1+Y)(2X+3)}{(1+X)^2}+\frac{(Y+2)X}{(1+X)}Q.
\end{align\*}


Now apply $C\_1$ to the linear term of $\Phi$ and apply $C\_2$ to the quartic term
in the manner $YQ^4 \le \operatorname{rhs}(C\_2)\,Q^2$. The result is exactly 0.


To show that $\Phi\gt 0$ strictly when $x,y,q\gt 0$, note that zero can only happen if $C\_1,C\_2$ hold with equality at the same time. But $Q\operatorname{rhs}(C\_1)-\operatorname{rhs}(C\_2)$ is manifestly negative.


