Inequality $abdc$ $\leq$ $3$ $a+b+c+d=6$
and
$a^2+b^2+c^2+d^2=12$.
and $a,b,c,d$ are reals.
Prove: $abcd$ $\leq$ $3$ without Lagrange multipliers, complex numbers or convexity help.
Using Cauchy–Schwarz inequality I found: $a,b,c,d \in [0,3]$.
How solve inequality?
 A: Another way, given you have already discovered $a, b, c, d \in [0, 3]$.  Note that the objective function $abcd$ and the constraints are symmetric and hence we may assume WLOG $0 \le a \le b \le c \le d \le 3$, or equivalently, $a \in [0, b], \; b \in [a, c], \; c \in [b, d], \; d \in [c, 3]$.  Further the objective is linear in each variable and the domain is closed and convex.  This means extrema can be attained only when each variable is at the boundary of its allowable interval.
If $a=0$, we clearly have a minimum for $abcd$, so $a=b$ at the maximum.  Similarly we must have $c\in \{b, d\}$ and $d \in \{c, 3\}$ .  Thus we have only two possibilities for the maximum, $(a, b, c, d) \in \{(p, p, q, q), (p, p, p, 3)\}$ for some $0< p \le q \le 3$.  
In the first case we have $p+q = 3, p^2+q^2=6$ and need to show $pq \le \sqrt3$, which follows from $2pq = (p+q)^2-(p^2+q^2) = 3 \le 2\sqrt3$ though the maximum is never reached in this case.
In the second case we have $2p+q = 3, 2p^2+q^2=3$ and need to show $p \le 1$ which is easy as the system has only one solution, $p=q=1$, with equality/ maximum attained.
A: Let $a=1+x$, $b=1+y$, $c=1+z$ and $d=1+t$.
Hence, $x+y+z+t=2$, $\sum\limits_{cyc}(1+x)^2=12$, which gives $x^2+y^2+z^2+t^2=4$ and
$xy+xz+yz+xt+yt+zt=0$.
By the way, $0=(xy+xz+yz+xt+yt+zt)^2=$
$=x^2y^2+x^2z^2+y^2z^2+x^2t^2+y^2t^2+z^2t^2+2\sum\limits_{cyc}x^2(yz+yt+zt)+6xyzt$.
Id est, $abcd=(1+x)(1+y)(1+z)(1+t)=3+\sum\limits_{cyc}xyz+xyzt=$
$=3+\frac{1}{2}\sum\limits_{cyc}x\sum\limits_{cyc}xyz+xyzt=3+\frac{1}{2}\left(\sum\limits_{cyc}x^2(yz+yt+xt)+4xyzt\right)+xyzt=$
$=3+\frac{1}{2}\sum\limits_{cyc}x^2(yz+yt+xt)+3xyzt=$
$=3+\frac{1}{4}\left(-(x^2y^2+x^2z^2+y^2z^2+x^2t^2+y^2t^2+z^2t^2)-6xyzt\right)+3xyzt=$
$=3-\frac{1}{4}(x^2y^2+x^2z^2+y^2z^2+x^2t^2+y^2t^2+z^2t^2)+\frac{3}{2}xyzt\leq$
$\leq3-\frac{3}{2}|xyzt|+\frac{3}{2}xyzt\leq3$.
A: If $abcd\le0$ there is nothing to prove, so either (1) $a,b>0$ and $c,d<0$ or (2) $a,b,c,d>0$.
Case (1). We have $a+b>6$, so $a^2+b^2\ge 3^2+3^2>12$. Contradiction.
Case (2) There is a standard result (well, standard amongst those who mess around with Olympiad inequalities) that for positive $a_1,\dots,a_n$ if we fix $\sum a_i$ and $\sum_{i\ne J}a_ia_j$, then the maximum of $a_1a_2\dots a_n$ is achieved when the $a_i$ assume at most two distinct values between them. So in this case we must have either (A) $a=b,c=d=3-a$ or (B) $b=c=d=6-a$.
Using $a^2+b^2+c^2+d^2$, (A) gives $ac=\frac{3}{2}$ and hence $abcd=\frac{9}{4}<3$, whilst (B) gives $a=b=c=1,d=3$ and hence $abcd=3$.
[see http://artofproblemsolving.com/community/c6h1187374p5777294 for the "generalised uvw" theorem]
A: Convexity methods or general inequalities help when the extremum is taken at a point where all variables have equal value. But in the problem at hand there is no feasible point with all $x_i$ equal. In other words: There is some "symmetry breaking", which then implies that there are several global maxima.
The set  $S$ of feasible points is a $2$-sphere embedded in ${\mathbb R}^4$, hence compact. Since the intersection between the hyperplane $x_1+x_2+x_3+x_4=6$ and the sphere $x_1^2+x_2^2+x_3^2+x_4^2=12$ is transversal the points where the objective function $p(x_1,x_2,x_3,x_4):=x_1x_2x_3x_4$ takes its maximum on $S$ will be brought to the fore using Lagrange's method. Set up the Lagrange function
$$\Phi(x_1,x_2,x_3,x_4):=p(x)-\lambda(x_1+x_2+x_3+x_4)-\mu(x_1^2+x_2^2+x_3^2+x_4^2)\ .$$
A conditionally stationary point $x_*=(x_1,x_2,x_3,x_4)$ of $p$ has to satisfy  the condition
$${\partial\Phi\over\partial x_1}(x_*)=x_2x_3x_4-\lambda-2\mu x_1=0\ ,$$
or
$$p(x_*)-\lambda x_1-2\mu x_1^2=0\ .$$
By symmetry this implies that all coordinates $x_i$ of  $x_*$  satisfy the same quadratic equation, hence there are at most two different coordinate values among the four. 
Trying $x_1=x_2=u$, $x_3=x_4=v$ leads to $$u+v=3,\quad u^2+v^2=6$$
and therefore $2uv=(u+v)^2-(u^2+v^2)=3$. From this we obtain $p(x_*)=u^2v^2={9\over4}<3$.
Trying $x_1=x_2=x_3=u$, $x_4=v$ leads to $$3u+v=6,\quad 3u^2+v^2=12\ ,$$
from which we obtain $(u,v)\in\{(1,3),(2,0)\}$. The corresponding $p$-values $u^3v$ are $3$ and $0$. It follows that the maximum of $p$ on the feasible set is $3$. 
Now that we have identified the extremal configurations we can start thinking about a proof not using Lagrange's method. A starting point could be the following: Show that any feasible point with $x_1<x_2<x_3$ cannot be optimal.
A: An altenative interpretation. By Vieta's formulas, we may assume that $a,b,c,d$ are four real roots of the polynomial
$$ q(t) = t^4-6t^3+12t^2+Kt+ p $$
and we have to prove that under such assumptions $p=abcd\geq 3$, so we have to study the cases for which
$$ f(t) = t(t-2)^3 =  t^4-6t^3+12t^3-8t = Ct-p $$
has four real solutions. That is the same as requesting that
$$ g(t) = \frac{1}{4}(2t^2-6t+3)^2 = t^4-6t^3+12t^3-9t+\frac{9}{4} = Dt-\left(p-\frac{9}{4}\right) $$
has four real solutions. Since $g(t)\geq 0$, that may happen only if $D\geq 0$. Moreover, that may happen only if the value of $Dt-\left(p-\frac{9}{4}\right)$ at $t=0$ is less than the value of $g(0)$. That gives $abcd\geq 0$, for starters. The midpoint of the roots of $g'(t)$ is at $t=\frac{3}{2}$, so we must have that the value of $Dt-\left(p-\frac{9}{4}\right)$ at $t=\frac{3}{2}$ is between $0$ and $g\left(\frac{3}{2}\right)=\frac{9}{16}$ (VRC-very restrictive constraint). Now it is enough to compute the equations of the tangent line to the graph of $g(t)$ at its first inflection point $t=1$ to get $\left(p-\frac{9}{4}\right)\leq \frac{3}{4}$, from which $\color{red}{abcd\leq 3}$ follows.

