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If x^3 + y^3 + z^3 = 81 then prove that x+y+z is less then or equal to 9 ?

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If x^3 + y^3 + z^3 = 81 then prove that x+y+z is less then or equal to 9 ?
posted Aug 22, 2016 by anonymous

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1 Answer

+1 vote

Thanks for the query.

As per C.S. Inequality -

(x.1 + y.1 + z.1)^2 < = (x^2 + y^2 + z^2)(1^2 + 1^2 + 1^2) = 3(x^2 + y^2 + z^2) ................... (1)

Again, applying C.S. Inequality on x^1/2, y^1/2 and z^1/2 we get,

(x^2 + y^2 + z^2)^2 < = (x^3 + y^3 + z^3)(x + y + z) = 81(x + y + z) .........................................(2)

Combining (1) and (2) we get,

(x + y + z)^4 < = 9(x^2 + y^2 + z^2)^2 which is again < = 9*81(x + y + z)

Therefore (x + y + z)^3 < = 9*81

or, x + y + z is less than equal to 9

Q.E.D.

answer Aug 23, 2016 by Indranil Datta
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