Question Number 120215 by benjo_mathlover last updated on 30/Oct/20
$${Prove}\:{that}\:\sqrt[{\mathrm{3}}]{\frac{\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)}{\mathrm{8}}\:}\:\geqslant\:\sqrt{\frac{{ab}+{bc}+{ca}}{\mathrm{3}}} \\ $$ $${for}\:{a},{b},{c}\:>\:\mathrm{0} \\ $$
Answered by TANMAY PANACEA last updated on 30/Oct/20
![((a+b)/2)≥(√(ab)) ((b+c)/2)≥(√(bc)) ((c+a)/2)≥(√(ac)) multiply them (((a+b)(b+c)(c+a))/8)≥(√(a^2 b^2 c^ )) (((a+b)(b+c)(c+a))/8)≥abc [(((a+b)(b+c)(c+a))/8)]^(1/3) ≥(abc)^(1/3) now(√((ab+bc+ac)/3)) ≥(√((ab×bc×ac)^(1/3) )) (√((ab+bc+ac)/3)) ≥(a^2 b^2 c^2 )^(1/6) (√((ab+bc+ac)/3)) ≥(abc)^(1/3)](Q120261.png)
$$\frac{{a}+{b}}{\mathrm{2}}\geqslant\sqrt{{ab}}\: \\ $$ $$\frac{{b}+{c}}{\mathrm{2}}\geqslant\sqrt{{bc}} \\ $$ $$\frac{{c}+{a}}{\mathrm{2}}\geqslant\sqrt{{ac}} \\ $$ $${multiply}\:{them} \\ $$ $$\frac{\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)}{\mathrm{8}}\geqslant\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{} } \\ $$ $$\frac{\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)}{\mathrm{8}}\geqslant{abc} \\ $$ $$\left[\frac{\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)}{\mathrm{8}}\right]^{\frac{\mathrm{1}}{\mathrm{3}}} \geqslant\left({abc}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$ $${now}\sqrt{\frac{{ab}+{bc}+{ac}}{\mathrm{3}}}\:\geqslant\sqrt{\left({ab}×{bc}×{ac}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} } \\ $$ $$\sqrt{\frac{{ab}+{bc}+{ac}}{\mathrm{3}}}\:\geqslant\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{6}}} \\ $$ $$\sqrt{\frac{{ab}+{bc}+{ac}}{\mathrm{3}}}\:\geqslant\left({abc}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$