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

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 ;)

$$\boldsymbol{{Lesson}}\mathrm{1}.\:\boldsymbol{\mathrm{AM}}−\boldsymbol{\mathrm{GM}}\:'\:\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{inequality}}\:\left(\boldsymbol{\mathrm{Cauchy}}\right) \\ $$ $$\boldsymbol{\mathrm{form}}\::\:\frac{\boldsymbol{{a}}_{\mathrm{1}} +\boldsymbol{{a}}_{\mathrm{2}} +...+\boldsymbol{{a}}_{\boldsymbol{{n}}} }{\boldsymbol{{n}}}\:\geqslant\:\sqrt[{\boldsymbol{{n}}}]{\boldsymbol{{a}}_{\mathrm{1}} \boldsymbol{{a}}_{\mathrm{2}} ...\boldsymbol{{a}}_{\boldsymbol{{n}}} } \\ $$ $$\boldsymbol{{where}}\:\boldsymbol{{a}}_{\mathrm{1}} ,\boldsymbol{{a}}_{\mathrm{2}} ,....,\boldsymbol{{a}}_{\boldsymbol{{n}}} >\mathrm{0} \\ $$ $$\boldsymbol{{E}\mathrm{qual}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{{a}}_{\mathrm{1}} =\boldsymbol{{a}}_{\mathrm{2}} =.....=\boldsymbol{{a}}_{\boldsymbol{{n}}} \\ $$ $$\boldsymbol{{e}}.\boldsymbol{{g}}.\:\mathrm{1}.\:{Given}\:{a},{b},{c}>\mathrm{0},\:{prove}\:{that}\: \\ $$ $$\:\:\:\:\:\:\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)\geqslant\mathrm{8}{abc} \\ $$ $$\boldsymbol{{Solu}}.\:{by}\:{AM}−{GM} \\ $$ $$\:\:\:\:\:\:\:\:{a}+{b}\:\geqslant\:\mathrm{2}\sqrt{{ab}}\:\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$ $$\:\:\:\:\:\:\:\:{b}+{c}\:\geqslant\:\mathrm{2}\sqrt{{bc}}\:\:\:\:\:\:\:\:\left(\mathrm{2}\right) \\ $$ $$\:\:\:\:\:\:\:\:{c}+{a}\:\geqslant\:\mathrm{2}\sqrt{{ca}}\:\:\:\:\:\:\:\left(\mathrm{3}\right) \\ $$ $$\:\left(\mathrm{1}\right)×\left(\mathrm{2}\right)×\left(\mathrm{3}\right)\:\Rightarrow\:\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)\geqslant\mathrm{8}\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} }=\mathrm{8}{abc} \\ $$ $$\boldsymbol{{Now}}\:\boldsymbol{{practice}}. \\ $$ $$\:.\:{Given}\:{a},{b},{c}>\mathrm{0}\:{prove}\:{that} \\ $$ $$\:\:\:\:\:\mathrm{1}.\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \geqslant{ab}+{bc}+{ca} \\ $$ $$\:\:\:\:\:\mathrm{2}.\:\left({a}+\frac{\mathrm{1}}{{b}}\right)\left({b}+\frac{\mathrm{1}}{{c}}\right)\left({c}+\frac{\mathrm{1}}{{a}}\right)\geqslant\mathrm{8} \\ $$ $$\:\:\:\:\:\mathrm{3}.\:\mathrm{4}\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right)\geqslant\left({a}+{b}\right)^{\mathrm{3}} \\ $$ $$\:\:\:\:\:\mathrm{4}.\:\:\mathrm{9}\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \right)\geqslant\left({a}+{b}+{c}\right)^{\mathrm{3}} \\ $$ $$\mathrm{let}'\mathrm{s}\:\mathrm{try},\:\mathrm{I}\:\mathrm{will}\:\mathrm{post}\:\mathrm{my}\:\mathrm{solution}\:\mathrm{for}\:\mathrm{which}\:\mathrm{one} \\ $$ $$\left.\mathrm{that}\:\mathrm{you}\:\mathrm{can}'\mathrm{t}\:\mathrm{do}\:;\right) \\ $$

Answered by Joel576 last updated on 03/Apr/17

(1) a + b ≥ 2(√(ab )) ⇔ a^2 + b^2 ≥ 2ab ... (i) b + c ≥ 2(√(bc)) ⇔ b^2 + c^2 ≥ 2bc ... (ii) a + c ≥ 2(√(ac)) ⇔ a^2 + c^2 ≥ 2ac ... (iii) (i) + (ii) + (iii) 2(a^2 + b^2 + c^2 ) ≥ 2(ab + bc + ac) ⇒ a^2 + b^2 + c^2 ≥ ab + bc + ac

$$\left(\mathrm{1}\right) \\ $$ $${a}\:+\:{b}\:\geqslant\:\mathrm{2}\sqrt{{ab}\:}\:\:\Leftrightarrow\:\:\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:\geqslant\:\mathrm{2}{ab}\:\:...\:\left({i}\right) \\ $$ $${b}\:+\:{c}\:\geqslant\:\mathrm{2}\sqrt{{bc}}\:\:\:\:\Leftrightarrow\:\:\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \:\geqslant\:\mathrm{2}{bc}\:\:\:...\:\left({ii}\right) \\ $$ $${a}\:+\:{c}\:\geqslant\:\mathrm{2}\sqrt{{ac}}\:\:\:\Leftrightarrow\:\:\:{a}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \:\geqslant\:\mathrm{2}{ac}\:\:...\:\left({iii}\right) \\ $$ $$ \\ $$ $$\left({i}\right)\:+\:\left({ii}\right)\:+\:\left({iii}\right) \\ $$ $$\mathrm{2}\left({a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \right)\:\geqslant\:\mathrm{2}\left({ab}\:+\:{bc}\:+\:{ac}\right) \\ $$ $$\Rightarrow\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \:\geqslant\:{ab}\:+\:{bc}\:+\:{ac} \\ $$

Answered by Joel576 last updated on 03/Apr/17

(2) (a + (1/b)) ≥ 2(√(a/b)) ... (i) (b + (1/c)) ≥ 2(√(b/c)) ... (ii) (c + (1/a)) ≥ 2(√(c/a)) ... (iii) (i) . (ii) . (iii) (a + (1/b))(b + (1/c))(c + (1/a)) ≥ 8(√1)

$$\left(\mathrm{2}\right) \\ $$ $$\left({a}\:+\:\frac{\mathrm{1}}{{b}}\right)\:\geqslant\:\mathrm{2}\sqrt{\frac{{a}}{{b}}}\:\:\:...\:\left({i}\right) \\ $$ $$\left({b}\:+\:\frac{\mathrm{1}}{{c}}\right)\:\geqslant\:\mathrm{2}\sqrt{\frac{{b}}{{c}}}\:\:\:...\:\left({ii}\right) \\ $$ $$\left({c}\:+\:\frac{\mathrm{1}}{{a}}\right)\:\geqslant\:\mathrm{2}\sqrt{\frac{{c}}{{a}}}\:\:\:...\:\left({iii}\right) \\ $$ $$ \\ $$ $$\left({i}\right)\:.\:\left({ii}\right)\:.\:\left({iii}\right) \\ $$ $$\left({a}\:+\:\frac{\mathrm{1}}{{b}}\right)\left({b}\:+\:\frac{\mathrm{1}}{{c}}\right)\left({c}\:+\:\frac{\mathrm{1}}{{a}}\right)\:\geqslant\:\mathrm{8}\sqrt{\mathrm{1}} \\ $$

Commented byMr Chheang Chantria last updated on 03/Apr/17

Very nice solution

$$\boldsymbol{{Very}}\:\boldsymbol{{nice}}\:\boldsymbol{{solution}} \\ $$

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