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 >0 Equal at a_1 =a_2 =.....=a_n e.g. 1. Given a,b,c>0, prove that (a+b)(b+c)(c+a)≥8abc Solu. by AM−GM a+b ≥ 2(√(ab)) (1) b+c ≥ 2(√(bc)) (2) c+a ≥ 2(√(ca)) (3) (1)×(2)×(3) ⇒ (a+b)(b+c)(c+a)≥8(√(a^2 b^2 c^2 ))=8abc Now practice. . Given a,b,c>0 prove that 1. a^2 +b^2 +c^2 ≥ab+bc+ca 2. (a+(1/b))(b+(1/c))(c+(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 ;)


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

$L e s s o n 1 . AM - GM^{'} s inequality (Cauchy)$ $form : \frac{a_{1} + a_{2} + . . . + a_{n}}{n} ⩾ \sqrt[n]{a_{1} a_{2} . . . a_{n}}$ $w h e r e a_{1}, a_{2}, . . . ., a_{n} > 0$ $E qual at a_{1} = a_{2} = . . . . . = a_{n}$ $e . g . 1 . G i v e n a, b, c > 0, p r o v e t h a t$ $(a + b) (b + c) (c + a) ⩾ 8 a b c$ $S o l u . b y A M - G M$ $a + b ⩾ 2 \sqrt{a b} (1)$ $b + c ⩾ 2 \sqrt{b c} (2)$ $c + a ⩾ 2 \sqrt{c a} (3)$ $(1) \times (2) \times (3) \Rightarrow (a + b) (b + c) (c + a) ⩾ 8 \sqrt{a^{2} b^{2} c^{2}} = 8 a b c$ $N o w p r a c t i c e .$ $. G i v e n a, b, c > 0 p r o v e t h a t$ $1 . a^{2} + b^{2} + c^{2} ⩾ a b + b c + c a$ $2 . (a + \frac{1}{b}) (b + \frac{1}{c}) (c + \frac{1}{a}) ⩾ 8$ $3 . 4 (a^{3} + b^{3}) ⩾ {(a + b)}^{3}$ $4 . 9 (a^{3} + b^{3} + c^{3}) ⩾ {(a + b + c)}^{3}$ ${let}^{'} s try, I will post my solution for which one$ $that you {can}^{'} t do;)$
	Answered by Joel576 last updated on 03/Apr/17

	$(1)$ $a + b ⩾ 2 \sqrt{a b} \Leftrightarrow a^{2} + b^{2} ⩾ 2 a b . . . (i)$ $b + c ⩾ 2 \sqrt{b c} \Leftrightarrow b^{2} + c^{2} ⩾ 2 b c . . . (i i)$ $a + c ⩾ 2 \sqrt{a c} \Leftrightarrow a^{2} + c^{2} ⩾ 2 a c . . . (i i i)$ $(i) + (i i) + (i i i)$ $2 (a^{2} + b^{2} + c^{2}) ⩾ 2 (a b + b c + a c)$ $\Rightarrow a^{2} + b^{2} + c^{2} ⩾ a b + b c + a c$
	Answered by Joel576 last updated on 03/Apr/17

	$(2)$ $(a + \frac{1}{b}) ⩾ 2 \sqrt{\frac{a}{b}} . . . (i)$ $(b + \frac{1}{c}) ⩾ 2 \sqrt{\frac{b}{c}} . . . (i i)$ $(c + \frac{1}{a}) ⩾ 2 \sqrt{\frac{c}{a}} . . . (i i i)$ $(i) . (i i) . (i i i)$ $(a + \frac{1}{b}) (b + \frac{1}{c}) (c + \frac{1}{a}) ⩾ 8 \sqrt{1}$
	Commented by Mr Chheang Chantria last updated on 03/Apr/17

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