Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 194808 by York12 last updated on 15/Jul/23

suppose a,b,c are positive real numbers prove the inequality (((a+b)/2))(((b+c)/2))(((c+a)/2))≥(((a+b+c)/3))(((abc)^2 ))^(1/3)

$$ \\ $$$${suppose}\:{a},{b},{c}\:{are}\:{positive}\:{real}\:{numbers} \\ $$$${prove}\:{the}\:{inequality} \\ $$$$\left(\frac{{a}+{b}}{\mathrm{2}}\right)\left(\frac{{b}+{c}}{\mathrm{2}}\right)\left(\frac{{c}+{a}}{\mathrm{2}}\right)\geqslant\left(\frac{{a}+{b}+{c}}{\mathrm{3}}\right)\sqrt[{\mathrm{3}}]{\left({abc}\right)^{\mathrm{2}} } \\ $$

Commented by York12 last updated on 16/Jul/23

guys , please help

$${guys}\:,\:{please}\:{help} \\ $$

Commented by sniper237 last updated on 16/Jul/23

call each member L and R Use arihmeico−geomeric inegaligy n=3 , ^3 (√((abc)^2 )) =^3 (√((ab)(ac)(bc))) ≤ ((ab+ac+bc)/3) recall (a+b)(a+c)(b+c)=(a+b+c)(ab+ac+bc)+abc L−R≥(((a+b)(a+c)(b+c)−8abc)/(8×9))≥0 cause for n=2, 2(√(ab)) ≤a+b ; 2(√(ac)) ≤a+c ; 2(√(bc)) ≤b+c

$${call}\:{each}\:{member}\:{L}\:{and}\:{R} \\ $$$${Use}\:{arihmeico}−{geomeric}\:{inegaligy} \\ $$$${n}=\mathrm{3}\:,\:\:^{\mathrm{3}} \sqrt{\left({abc}\right)^{\mathrm{2}} \:}\:=\:^{\mathrm{3}} \sqrt{\left({ab}\right)\left({ac}\right)\left({bc}\right)}\:\leqslant\:\frac{{ab}+{ac}+{bc}}{\mathrm{3}} \\ $$$${recall}\:\left({a}+{b}\right)\left({a}+{c}\right)\left({b}+{c}\right)=\left({a}+{b}+{c}\right)\left({ab}+{ac}+{bc}\right)+{abc} \\ $$$$\:{L}−{R}\geqslant\frac{\left({a}+{b}\right)\left({a}+{c}\right)\left({b}+{c}\right)−\mathrm{8}{abc}}{\mathrm{8}×\mathrm{9}}\geqslant\mathrm{0}\:{cause}\: \\ $$$${for}\:\:{n}=\mathrm{2},\:\:\mathrm{2}\sqrt{{ab}}\:\leqslant{a}+{b}\:;\:\mathrm{2}\sqrt{{ac}}\:\leqslant{a}+{c}\:;\:\mathrm{2}\sqrt{{bc}}\:\leqslant{b}+{c} \\ $$$$ \\ $$

Commented by York12 last updated on 16/Jul/23

thanks sir

$${thanks}\:{sir}\: \\ $$

Commented by York12 last updated on 17/Jul/23

(((a+b)/2))(((b+c)/2))(((a+c)/2))−(((a+b+c)/3))(((abc)^2 ))^(1/3) ≥ (((a+b)/2))(((b+c)/2))(((a+c)/2))_((((a+b+c)(ab+bc+ac))/9)) =(((a+b)(b+c)(a+c))/8)−(((a+b)(b+c)(a+c)+abc)/9) =(a+b)(b+c)(a+c)((1/8)−(1/9))−((abc)/9)≥((abc)/9)−((abc)/9)=0 ⇒(((a+b)/2))(((b+c)/2))(((a+c)/2))−(((a+b+c)/3))(((abc)^2 ))^(1/3) ≥0 ⇒(((a+b)/2))(((b+c)/2))(((a+c)/2))≥(((a+b+c)/3))(((abc)^2 ))^(1/3) →(Hence proved)

$$\left(\frac{{a}+{b}}{\mathrm{2}}\right)\left(\frac{{b}+{c}}{\mathrm{2}}\right)\left(\frac{{a}+{c}}{\mathrm{2}}\right)−\left(\frac{{a}+{b}+{c}}{\mathrm{3}}\right)\sqrt[{\mathrm{3}}]{\left({abc}\right)^{\mathrm{2}} }\geqslant \\ $$$$\left(\frac{{a}+{b}}{\mathrm{2}}\right)\left(\frac{{b}+{c}}{\mathrm{2}}\right)\left(\frac{{a}+{c}}{\mathrm{2}}\right)\_\left(\frac{\left({a}+{b}+{c}\right)\left({ab}+{bc}+{ac}\right)}{\mathrm{9}}\right) \\ $$$$=\frac{\left({a}+{b}\right)\left({b}+{c}\right)\left({a}+{c}\right)}{\mathrm{8}}−\frac{\left({a}+{b}\right)\left({b}+{c}\right)\left({a}+{c}\right)+{abc}}{\mathrm{9}} \\ $$$$=\left({a}+{b}\right)\left({b}+{c}\right)\left({a}+{c}\right)\left(\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{9}}\right)−\frac{{abc}}{\mathrm{9}}\geqslant\frac{{abc}}{\mathrm{9}}−\frac{{abc}}{\mathrm{9}}=\mathrm{0} \\ $$$$\Rightarrow\left(\frac{{a}+{b}}{\mathrm{2}}\right)\left(\frac{{b}+{c}}{\mathrm{2}}\right)\left(\frac{{a}+{c}}{\mathrm{2}}\right)−\left(\frac{{a}+{b}+{c}}{\mathrm{3}}\right)\sqrt[{\mathrm{3}}]{\left({abc}\right)^{\mathrm{2}} }\geqslant\mathrm{0} \\ $$$$\Rightarrow\left(\frac{{a}+{b}}{\mathrm{2}}\right)\left(\frac{{b}+{c}}{\mathrm{2}}\right)\left(\frac{{a}+{c}}{\mathrm{2}}\right)\geqslant\left(\frac{{a}+{b}+{c}}{\mathrm{3}}\right)\sqrt[{\mathrm{3}}]{\left({abc}\right)^{\mathrm{2}} }\rightarrow\left({Hence}\:{proved}\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com