Question Number 6154 by Rasheed Soomro last updated on 16/Jun/16
$${Prove}\:{or}\:{disprove} \\ $$$$\left(\frac{\mathrm{2}\boldsymbol{{ab}}+\mathrm{2}\boldsymbol{{bc}}+\boldsymbol{{ca}}}{\mathrm{5}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \geqslant\:\:\boldsymbol{{abc}}\:\:\:\forall\:\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}}>\mathrm{0} \\ $$
Commented by Yozzii last updated on 18/Jun/16
$$\frac{\mathrm{5}\sqrt{\mathrm{5}}}{\left(\frac{\mathrm{2}}{{c}}+\frac{\mathrm{2}}{{a}}+\frac{\mathrm{1}}{{b}}\right)\sqrt{\mathrm{2}{ab}+\mathrm{2}{bc}+{ca}}}\leqslant\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{b}}+\frac{\mathrm{2}}{{a}}+\frac{\mathrm{2}}{{c}}\geqslant\mathrm{5}\sqrt{\frac{\mathrm{5}}{\mathrm{2}{ab}+\mathrm{2}{bc}+{ca}}} \\ $$$$\left(\frac{\mathrm{1}}{{b}}+\frac{\mathrm{2}}{{a}}+\frac{\mathrm{2}}{{c}}\right)^{\mathrm{2}} \geqslant\frac{\mathrm{125}}{\mathrm{2}{ab}+\mathrm{2}{bc}+{ca}} \\ $$$$\left(\frac{\mathrm{1}}{{b}}+\frac{\mathrm{2}}{{a}}+\frac{\mathrm{2}}{{c}}\right)^{\mathrm{3}} \geqslant\frac{\mathrm{125}}{{abc}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{{b}}+\frac{\mathrm{2}}{{a}}+\frac{\mathrm{2}}{{c}}\right)\geqslant\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{125}}{{abc}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{{b}}+\frac{\mathrm{2}}{{a}}+\frac{\mathrm{2}}{{c}}\right)\geqslant\left(\mathrm{1}.\mathrm{666}…\right).\frac{\mathrm{1}}{\left({abc}\right)^{\mathrm{1}/\mathrm{3}} } \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${By}\:{AM}−{GM}, \\ $$$$\frac{{b}^{−\mathrm{1}} +\mathrm{2}{a}^{−\mathrm{1}} +\mathrm{2}{c}^{−\mathrm{1}} }{\mathrm{3}}\geqslant\left({b}^{−\mathrm{1}} ×\mathrm{2}{a}^{−\mathrm{1}} ×\mathrm{2}{c}^{−\mathrm{1}} \right)^{\mathrm{1}/\mathrm{3}} \\ $$$$ \\ $$