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?

Solution 1:

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.

Solution 2:

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, $abcd=(1+x)(1+y)(1+z)(1+t)=3+\sum\limits_{cyc}xyz+xyzt=$

$=3+\frac{1}{2}\sum\limits_{cyc}x\sum\limits_{cyc}xyz+xyzt=3+\frac{1}{2}\left(\sum\limits_{cyc}x^2(yz+yt+xt)+4xyzt\right)+xyzt=$

$=3+\frac{1}{2}\sum\limits_{cyc}x^2(yz+yt+xt)+3xyzt=$

$=3+\frac{1}{4}\left(-(x^2y^2+x^2z^2+y^2z^2+x^2t^2+y^2t^2+z^2t^2)-6xyzt\right)+3xyzt=$

$=3-\frac{1}{4}(x^2y^2+x^2z^2+y^2z^2+x^2t^2+y^2t^2+z^2t^2)+\frac{3}{2}xyzt\leq$

$\leq3-\frac{3}{2}|xyzt|+\frac{3}{2}xyzt\leq3$.