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Prove that the greatest value of `x y` is `c^3//sqrt(2a b)dot` if `a^2x^4+b^4y^4=c^6dot`

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Verified by Experts

Using `A.M ge G.M`., we get
`(a^2x^4+b^2y^4)/(2) ge 9a^2x^4b^2y^4)^(1)/(2)`
`therefore (e^6)/(2) ge (a^2x^4b^2y^4)^(1)/(2)`
`rArr (c^6)/(2) ge abx^2y^2`
`rArr xyle (c^3)/(sqrt(2ab))`
Hence , `(xy)_("max")=(c^3)/(sqrt(2ab))`
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