<s>[INST] Find the minimum value of this expression. Let $x$ and $y$ be two real numbers satisfying $$(x+y)^3 +4xy \geqslant 2.$$
Find the minimum value of the expression
$$E=x^3+y^3 -2(x^2+y^2)-1.$$
I tried. We have
$$2\leqslant (x+y)^3+(x+y)^2.$$
Therefore, $$x+y \geqslant 1.$$
$$E = (x+y)^3-3xy(x+y)-2[(x+y)^2-2xy]-1.$$
Or
$$E = (x+y)^3-xy(3x+3y-4)-2(x+y)^2-1.$$
Using $$xy\leqslant \dfrac{(x+y)^2}{4}$$ and if $$3x + 3y -4 >0,$$
$$E \geqslant (x+y)^3-\dfrac{(x+y)^2}{4}(3x+3y-4).$$
Put $t = x + y$, and using derivative.
A problem is, if $$3x + 3y -4 <0,$$ how to prove?
 [/INST] when $x,y$ are non negative. the case is very easy as $x^3-2x^2$ have min value when $x>0,\to E_{min}=-\dfrac{91}{27}$ .
if $x \le 0,y>0$, let $x=-u, u \ge 0$, let $y=3u+a,$, it is not difficult to prove 
when $a \ge 1.3, (2u+a)^3-4u(3u+a)\ge 2$ 
$E=26u^3+(27a-20)u^2+(9a^2-12a)u+a^3-2a^2-1$
it is obvious that $E_{min}=a^3-2a^2-1$ when $u=0 \to E_{min}=-\dfrac{59}{27}$
so final answer is when $x=y=\dfrac{4}{3},$ min is $-\dfrac{91}{27}$
edit: I add the reason why $x,y$ can't be negative at same time.
if $x,y$ are both negative, let $x=-u, y=-v, \implies -(u+v)^3+4uv \ge 2 \iff (u+v)^3-4uv+2 \le 0$  with $u>0,v>0$
but $(u+v)^3 \ge (2\sqrt{uv})^3$, let $t=\sqrt{uv}$
LHS$\ge 8t^3-4t^2+2=(2t+1)((t-1)^2+1) > 0$ so $x,y$ can't be negative at same time.
edit 3: indeed, the last step is not necessary as op already show $x+y \ge 1$ which hints $x,y$ can't be negative at same time.
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