Want to prove an inequality 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.
 A: Welcome to MO! However, your conjecture is false e.g. for $q = 1/2$, $x= 3/8$, $y = 1/4$, $t= 0$. 
Added: The OP later stated that the additional condition $t=3$ was initially omitted in the OP's post. Anyhow, the problem is one of real algebraic geometry and, as such, admits a completely algorithmic solution. In Mathematica, such algorithms are represented by Reduce[] and related commands. Using Reduce[] indeed, we get 

This proves the conjecture. We see that it took Mathematica about 2 sec to obtain this result; so, a manual proof might be quite long and laborious, and most likely less reliable than Mathematica's.  
A: Iosif's answer 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.
A: Remark: Actually, the inequality is quite easy. The trick is to take $x, y$ as parameters.
Denote the expression by $f(q)$.
It suffices to prove that $f(q) \ge 0$ provided that
\begin{align*}
    &x, y > 0,\\
    &x + y < 1,\\
    &3y > (x + y)^2,\\
    &3x > (x + y)^2, \\
    &y < q < 1 - x, \\
    &1 + y - \frac{3y}{x + y} \le q \le \frac{3x}{x + y} - x.
\end{align*}
(Note: Condition $(x+y)\left[1+max\{\frac{1-q}{y},\frac{q}{x}\}\right]\leq3$ is equivalent to $(x + y)[1 + (1-q)/y] \le 3$
and $(x + y)(1 + q/x) \le 3$ which are equivalent to $1 + y - \frac{3y}{x + y} \le q \le \frac{3x}{x + y} - x$.
Also, from $y < \frac{3x}{x + y} - x$, we have $3x > (x + y)^2$,
and from $1 + y - \frac{3y}{x + y} < 1 - x $, we have $3y > (x + y)^2$.)
Note that $f(q)$ is quadratic and concave (the coefficient of $q^2$
is $-3x - 3y$).
Also, we have
\begin{align*}
    f(y) &= \frac{3x(1 - x - y)(2y - x)}{x + y},\\
    f(1 - x) &= \frac{3y(1 - x - y)(2x - y)}{x + y},\\
    f\left(1 + y - \frac{3y}{x + y}\right) &= \frac{3y(2 - x - y)(x - 2y)}{x + y}, \\
    f\left(\frac{3x}{x + y} - x\right) &= \frac{3x(2 - x - y)(y - 2x)}{x + y}.
\end{align*}
We split into three cases:
(1) If $x > 2y$, we have $1 + y - \frac{3y}{x + y} > y$
and $1 - x \le \frac{3x}{x + y} - x$, and
$f(1 + y - \frac{3y}{x + y}) \ge 0$ and $f(1 - x) \ge 0$. Thus, $f(q) \ge 0$.
(2) If $y/2 \le x \le 2y$, we have $1 + y - \frac{3y}{x + y} \le y$
and $1 - x \le \frac{3x}{x + y} - x$, and
$f(y) \ge 0$ and $f(1 - x) \ge 0$. Thus, we have $f(q) \ge 0$.
(3) If $x < y/2$, we have $1 + y - \frac{3y}{x + y} \le y$
and $1 - x > \frac{3x}{x + y} - x$, and
$f(y) \ge 0$ and $f(\frac{3x}{x + y} - x) \ge 0$. Thus, $f(q) \ge 0$.
We are done.
