Menu Close

Prove-or-disprove-2ab-2bc-ca-5-3-2-abc-a-b-c-gt-0-




Question Number 6154 by Rasheed Soomro last updated on 16/Jun/16
Prove or disprove  (((2ab+2bc+ca)/5))^(3/2) ≥  abc   ∀ a,b,c>0
$${Prove}\:{or}\:{disprove} \\ $$$$\left(\frac{\mathrm{2}\boldsymbol{{ab}}+\mathrm{2}\boldsymbol{{bc}}+\boldsymbol{{ca}}}{\mathrm{5}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \geqslant\:\:\boldsymbol{{abc}}\:\:\:\forall\:\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}}>\mathrm{0} \\ $$
Commented by Yozzii last updated on 18/Jun/16
((5(√5))/(((2/c)+(2/a)+(1/b))(√(2ab+2bc+ca))))≤1  (1/b)+(2/a)+(2/c)≥5(√(5/(2ab+2bc+ca)))  ((1/b)+(2/a)+(2/c))^2 ≥((125)/(2ab+2bc+ca))  ((1/b)+(2/a)+(2/c))^3 ≥((125)/(abc))  (1/3)((1/b)+(2/a)+(2/c))≥(1/3)(((125)/(abc)))^(1/3)   (1/3)((1/b)+(2/a)+(2/c))≥(1.666...).(1/((abc)^(1/3) ))  −−−−−−−−−−−−−−−−−−−−−−−−  By AM−GM,  ((b^(−1) +2a^(−1) +2c^(−1) )/3)≥(b^(−1) ×2a^(−1) ×2c^(−1) )^(1/3)
$$\frac{\mathrm{5}\sqrt{\mathrm{5}}}{\left(\frac{\mathrm{2}}{{c}}+\frac{\mathrm{2}}{{a}}+\frac{\mathrm{1}}{{b}}\right)\sqrt{\mathrm{2}{ab}+\mathrm{2}{bc}+{ca}}}\leqslant\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{b}}+\frac{\mathrm{2}}{{a}}+\frac{\mathrm{2}}{{c}}\geqslant\mathrm{5}\sqrt{\frac{\mathrm{5}}{\mathrm{2}{ab}+\mathrm{2}{bc}+{ca}}} \\ $$$$\left(\frac{\mathrm{1}}{{b}}+\frac{\mathrm{2}}{{a}}+\frac{\mathrm{2}}{{c}}\right)^{\mathrm{2}} \geqslant\frac{\mathrm{125}}{\mathrm{2}{ab}+\mathrm{2}{bc}+{ca}} \\ $$$$\left(\frac{\mathrm{1}}{{b}}+\frac{\mathrm{2}}{{a}}+\frac{\mathrm{2}}{{c}}\right)^{\mathrm{3}} \geqslant\frac{\mathrm{125}}{{abc}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{{b}}+\frac{\mathrm{2}}{{a}}+\frac{\mathrm{2}}{{c}}\right)\geqslant\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{125}}{{abc}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{{b}}+\frac{\mathrm{2}}{{a}}+\frac{\mathrm{2}}{{c}}\right)\geqslant\left(\mathrm{1}.\mathrm{666}…\right).\frac{\mathrm{1}}{\left({abc}\right)^{\mathrm{1}/\mathrm{3}} } \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${By}\:{AM}−{GM}, \\ $$$$\frac{{b}^{−\mathrm{1}} +\mathrm{2}{a}^{−\mathrm{1}} +\mathrm{2}{c}^{−\mathrm{1}} }{\mathrm{3}}\geqslant\left({b}^{−\mathrm{1}} ×\mathrm{2}{a}^{−\mathrm{1}} ×\mathrm{2}{c}^{−\mathrm{1}} \right)^{\mathrm{1}/\mathrm{3}} \\ $$$$ \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *