Misalkan a, b, c bilangan real positif yang memenuhi a + b + c = 1. Nilai minimum dari[tex] \dfrac{a.... Question from @iwansidh

MrAZA01

misalkan

a + b = x

a + b + c = 1 → x + c = 1

berlaku ketidaksamaan

AM ≥ GM

(x + c)/2 ≥ √xc

1/2 ≥ √xc

1/4 ≥ xc

tanda persamaan saat x = c,

maka didapat x = c = √1/4 = 1/2

x = a + b

a + b = 1/2

AM ≥ GM

(a + b)/2 ≥ √ab

1/4 ≥ √ab

1/16 ≥ ab

a = b, maka a = b = √1/16 = 1/4

nilai min

(a + b)/(abc)

= (1/4 + 1/4)/(1/4 . 1/4 . 1/2)

= (1/2)/(1/32)

= 32/2

= 16

iwansidh Pertidaksamaan AM ≥ GM itu maksudny apa y Kak, dapat ny dari mana?

MrAZA01 di komentar aja ka

MrAZA01 ga muat jelasin disini

MrAZA01 eh dm mksdny

iwansidh dm di slack jg bisa

nursahel26

$rh \: \leqslant rg \leqslant ra \leqslant rk$

$\frac{ 3 }{ \frac{1}{a} + \frac{1}{b} } \leqslant \sqrt[3]{abc} \leqslant \frac{a + b + c}{3} \leqslant \sqrt{ \frac{ {a}^{2} + {b}^{2} + {c}^{2} }{3} } \\$

$\frac{a + b}{abc} = \frac{a}{abc} + \frac{b}{abc} = \frac{1}{bc} + \frac{1}{ac} \\$

a + b + c = 1

$RA \geqslant RH$

$\frac{p + q}{2} \geqslant \frac{2}{ \frac{1}{p} + \frac{1}{q} } \\$

$\frac{1}{p} + \frac{1}{q} \geqslant \frac{4}{p + q} \\$

$\frac{1}{bc} + \frac{1}{ac} \geqslant \frac{4}{bc + ac} \\$

$\frac{a + b}{abc} = \frac{4}{c(b + a)} = \frac{4}{c(1 - c)} \\$

$= \frac{4}{c - {c}^{2} } \\$

$( \frac{a + b}{abc})Minimum \: = \frac{4}{Maximum} = \frac{4}{ \frac{d}{ - 4a} } \\$

$= \frac{4}{ \frac{ {1}^{2} - 4( - 1) - 0 }{ - 4( - 1)} } = \frac{4}{ \frac{1}{4} } = 16\\$

iwansidh Rumus rh ≤ rg ≤ ra ≤ rk menu dari mana?

nursahel26 Itu tuh ada rumusnya dah pokoknya

No 27. Mohon di jawab dengan penjelasan yang sederhana dan mudah di pahami. Trim's.

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