<s>[INST] Maximization Of Multivariable function let $F(a,b,c,d,e,f) = abcdef$ and our constraint is  $a^2+b^2+c^2+d^2+e^2+f^2=6$ ,$a+b+c+d+e+f=0$ then lagrange
we get
$abcdef = 2a^2k +ma=2b^2.k+m.b =.....=2f^2.k+mf$ but this is quite complicated to do is there any elegant way like trig sub or something
for $a,b,c,d,e,f$ real numbers
 [/INST] The following satisfy the constraints:
$$a=b=c=d=\frac{1}{\sqrt2}$$
$$e=f=-\frac{2}{\sqrt2}$$
$$abcdef=\frac12$$
In fact, this will turn out to be the maximum, but for now, we just need to note that  the maximum is greater than zero.
$a+b+c+d+e+f=0$ tells us that some are positive and some are negative. In fact, the number of positive values must be even because $abcdef>0$.
We assume that there are 4 positive values (a,b,c,d) and two negative (e,f). The proof for 2 positive and 4 negative is much the same.
Define
$$E=-e$$
$$F=-f$$
$$a+b+c+d=E+F=S$$
$$E^2+F^2=Q$$
$$a^2+b^2+c^2+d^2=6-Q$$
Now apply the inequality of geometric, arithmetic and quadratic means:
$$(abcd)^{\frac14} \leqslant \frac{S}{4}  \leqslant \left(\frac{6-Q}{4}\right)^{\frac12} \tag1$$
$$(EF)^{\frac12} \leqslant \frac{S}{2}  \leqslant \left(\frac{Q}{2}\right)^{\frac12} \tag2$$
From the second parts of (1) and (2):
$$S^2 \leqslant 24-4Q$$
$$S^2 \leqslant 2Q$$
and combining these gives
$$S\leqslant 2 \sqrt 2 \tag3 $$
From the first parts of (1) and (2):
$$abcd \leqslant \frac{S^4}{256}$$
$$ef=EF \leqslant \frac{S^2}{4}$$
Multiply:
$$abcdef \leqslant \frac{S^6}{1024} \leqslant  \frac{( 2 \sqrt 2 )^6}{1024}=\frac12 $$
As required.
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