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?

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.

