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

1 votes Thanks 2

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

$RA \geqslant RH$

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

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

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

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

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

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

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

2 votes Thanks 2

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.

Recommend Questions

Life Enjoy

Get in touch

Social