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Now, it is necessary to find


Is this possible and how? a,b,c are given constants.

I think, ? is probably a complicated function $F=F(a,b,c)$

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up vote 4 down vote accepted

Without loss of generality we can assume $a\ge b\ge c>0$.

Then $$\frac{x^2}{a^4}+\frac{y^2}{a^2b^2}+\frac{z^2}{a^2c^2}\le\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}\le\frac{x^2}{a^2c^2}+\frac{y^2}{b^2c^2}+\frac{z^2}{c^4}.$$

Noting that $$\frac{x^2}{a^4}+\frac{y^2}{a^2b^2}+\frac{z^2}{a^2c^2} =\frac1{a^2}\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)$$ $$\frac{x^2}{a^2c^2}+\frac{y^2}{b^2c^2}+\frac{z^2}{c^4}=\frac1{c^2}\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right),$$

we have $$\frac1{a^2}\le \frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}\le \frac1{c^2},$$

where equality holds, for example, for $(x,y,z)=(a,0,0)$ and $(x,y,z)=(0,0,c)$. (There are other sets of values $(x,y,z)$ for which the equality holds).

More generally (i.e. if we do not assume $a\ge b\ge c>0$), $$\frac1{\max(|a|,|b|,|c|)^2}\le \frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}\le \frac1{\min(|a|,|b|,|c|)^2}.$$

Edit. We can further show that, when $a\ge b\ge c>0$, $\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}$ can take any value $k\in \left[\frac1{a^2}, \frac1{c^2}\right]$. To show this we can consider $\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}$ (where $(x,y,z)=(at,\ 0,\ c(1-t))$) as a continuous function of $t\in [0,1]$ and use the intermediate value theorem.

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This is really an enlightening answer to my darkness! Thanks! – Martin Gales Nov 10 '11 at 6:40

The expression $$\frac{x^2}{a^4} + \frac{y^2}{b^4} + \frac{z^2}{c^4}$$ isn't constant on the ellipsoid $$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1.$$ Notice that I get different values on the points $(x,y,z) = (a,0,0)$ and $(x,y,z)=(0,b,0)$.

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Another Brown student on the site! My gosh (David Lowry). – mixedmath Nov 9 '11 at 6:04
How is the above observation relevant to the question? – tards Nov 9 '11 at 6:04
@tards: OP wants to find $?$, but the observation is that the constraint does not determine $?$ uniquely. Do you not see the relevance of this? – anon Nov 9 '11 at 6:08
@anon well $a^{-2} + b^{-2}$ satisfies the values at the two points. Perhaps, some function of $a, b, c$ will work? I am sorry I do not see the relevance. – tards Nov 9 '11 at 6:14
@tards: No, it doesn't; you get $a^{-2}$ in the first point, not $a^{-2}+b^{-2}$; you get $b^{-2}$ in the second point, not $a^{-2}+b^{-2}$; if $a\neq b$, then the two answers are different; the OP wants an expression in terms of $a$, $b$, and $c$; but if $a$, $b$, and $c$ are constant, then the expression will be itself constant. We cannot have a constant expression take two different values. Of course, it would be nice to have another pair of points in the case $a=b=c$. – Arturo Magidin Nov 9 '11 at 6:17

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