<s>[INST] 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?
 [/INST] 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]
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