Lesson1-AM-GM-s-inequality-Cauchy-form-a-1-a-2-a-n-n-a-1-a-2-a-n-1-n-where-a-1-a-2-a-n-gt-0-Equal-at-a-1-a-2-a-n-e-g-1-Given-a-b-c-gt-0-prove-that-

Question Number 11862 by Mr Chheang Chantria last updated on 03/Apr/17

L e s s o n 1 . AM - GM^{'} s inequality (Cauchy)

form : \frac{a_{1} + a_{2} + \dots + a_{n}}{n} ⩾ \sqrt[n]{a_{1} a_{2} \dots a_{n}}

w h e r e a_{1}, a_{2}, \dots ., a_{n} > 0

E qual at a_{1} = a_{2} = \dots . . = a_{n}

e . g . 1 . G i v e n a, b, c > 0, p r o v e t h a t

(a + b) (b + c) (c + a) ⩾ 8 a b c

S o l u . b y A M - G M

a + b ⩾ 2 \sqrt{a b} (1)

b + c ⩾ 2 \sqrt{b c} (2)

c + a ⩾ 2 \sqrt{c a} (3)

(1) \times (2) \times (3) \Rightarrow (a + b) (b + c) (c + a) ⩾ 8 \sqrt{a^{2} b^{2} c^{2}} = 8 a b c

N o w p r a c t i c e .

. G i v e n a, b, c > 0 p r o v e t h a t

1 . a^{2} + b^{2} + c^{2} ⩾ a b + b c + c a

2 . (a + \frac{1}{b}) (b + \frac{1}{c}) (c + \frac{1}{a}) ⩾ 8

3 . 4 (a^{3} + b^{3}) ⩾ {(a + b)}^{3}

4 . 9 (a^{3} + b^{3} + c^{3}) ⩾ {(a + b + c)}^{3}

{let}^{'} s try, I will post my solution for which one

that you {can}^{'} t do;)

Answered by Joel576 last updated on 03/Apr/17

(1)

a + b ⩾ 2 \sqrt{a b} \Leftrightarrow a^{2} + b^{2} ⩾ 2 a b \dots (i)

b + c ⩾ 2 \sqrt{b c} \Leftrightarrow b^{2} + c^{2} ⩾ 2 b c \dots (i i)

a + c ⩾ 2 \sqrt{a c} \Leftrightarrow a^{2} + c^{2} ⩾ 2 a c \dots (i i i)

(i) + (i i) + (i i i)

2 (a^{2} + b^{2} + c^{2}) ⩾ 2 (a b + b c + a c)

\Rightarrow a^{2} + b^{2} + c^{2} ⩾ a b + b c + a c

Answered by Joel576 last updated on 03/Apr/17

(2)

(a + \frac{1}{b}) ⩾ 2 \sqrt{\frac{a}{b}} \dots (i)

(b + \frac{1}{c}) ⩾ 2 \sqrt{\frac{b}{c}} \dots (i i)

(c + \frac{1}{a}) ⩾ 2 \sqrt{\frac{c}{a}} \dots (i i i)

(i) . (i i) . (i i i)

(a + \frac{1}{b}) (b + \frac{1}{c}) (c + \frac{1}{a}) ⩾ 8 \sqrt{1}

Commented by Mr Chheang Chantria last updated on 03/Apr/17

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