Suppose that x, y, z are positive numbers and 4x + y + 9z = 11, find the minimum

Question

Suppose that x, y, z are positive numbers and 4x + y + 9z = 11, find the minimum value of

1/x + 9/y +4/z

an answer given to me is below but I have a few questions on the answer below THANKS! :)

(4x+y+9z)^2(1/x+9/y+4/z)^2>=(4x*1/x+y*9/y+9z*4/z)^2=(4+9+36)^2

(4x+y+9z)^2*11^2>=49^2
(4x+y+9z)^2>=49^2/11^2

so the minimum is 49/11?


can anyone explain how we get 11^2 in the part of the equation ---> (4x+y+9z)^2*11^2>=49^2 or is it incorrect?

did it replace this part of the equation (1/x+9/y+4/z)^2?

Explanation / Answer

it is wrong because first you have asked for minimum value of 1/x + 9/y +4/z but he give you minimum of 4x + y + 9z = 11 second he don't follow any mathematical rule of equality so this is wrong

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