I have $3$ vectors: $$A =(a_1, a_2, a_3), B=(b_1, b_2, b_3), C= (c_1, c_2, c_3)$$ I want to find maximum of triple product of these vectors, with a condition that: $$ (a_1^2+a_2^2+a_3^2) +(b_1^2+b_2^2+b_3^2)+(c_1^2+c_2^2+c_3^2)\leq 1$$
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This is maximising the volume of a parallelepiped spanned by vectors $\bf a$, $\bf b$ and $\bf c$ under the condition that $|\mathbf{a}|^2+ |\mathbf{b}|^2+|\mathbf{c}|^2\le1$. If we insist $|\mathbf{a}|=r$, $|\mathbf{b}|=s$ and $|\mathbf{c}|=t$ for fixed $r$, $s$ and $t$ then the volume is maximised at a cuboid, of volume $rst$. So we have to maximise $rst$ under the condition $r^2+s^2+t^2=1$. By AM-GM then $(rst)^{2/3}\le1/3$ so $rst\le 3^{-3/2}$. So that's the maximum acheived when $r=s=t=3^{-1/2}$.
answered Apr 23, 2017 at 11:47
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$\begingroup$ i understand that, but can you please extend the proof that maximum can be achieved at cuboid? $\endgroup$– llamaApr 23, 2017 at 11:56
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$\begingroup$ @llama This is nothing more than the observation that $|u\cdot(v\times w)|\le|u||v\times w|\le |u||v||w|$. $\endgroup$ Apr 23, 2017 at 12:09