Problem with inequality

Algebra Level 3

If x , y , z x,y,z are positive real such that x + y + z = 6 x+y+z=6 and x 3 + y 3 + z 3 = 24 x^3+y^3+z^3=24 . Find 6 x + 7 y 2 + 8 z 3 6x+7y^2+8z^3 .


The answer is 104.

Gramos Nikci by Gramos Nikci , 20, Serbia

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1 solution

Steven Yuan
Jul 11, 2017

By Hölder's Inequality,

[ ( x 3 + y 3 + z 3 ) ( 1 + 1 + 1 ) ( 1 + 1 + 1 ) ] 1 / 3 ≥ x + y + z , [(x^3 + y^3 + z^3)(1 + 1 + 1)(1 + 1 + 1)]^{1/3} \geq x + y + z,

or

9 ( x 3 + y 3 + z 3 ) ≥ ( x + y + z ) 3 , 9(x^3 + y^3 + z^3) \geq (x + y + z)^3,

with equality occurring when x = y = z . x = y = z. Plugging in our given values yields 9 ( 24 ) ≥ 6 3 , 9(24) \geq 6^3, which is actually an equality. Thus, we see that x = y = z , x = y = z, which means they all must be equal to 2. Therefore, 6 x + 7 y 2 + 8 z 3 = 6 ( 2 ) + 7 ( 2 2 ) + 8 ( 2 3 ) = 104 . 6x + 7y^2 + 8z^3 = 6(2) + 7(2^2) + 8(2^3) = \boxed{104}.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

power mean inequality (qagh) is more natural (in my opinion).

Pi Han Goh - 3 years, 11 months ago

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