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

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