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Question Number 55326 by tm888 last updated on 21/Feb/19
find  the minimum value  (a/( (√(a^2 +8bc))))+(b/( (√(b^2 +8ac))))+(c/( (√(c^2 +8ab))))
$${find}\:\:{the}\:{minimum}\:{value} \\ $$$$\frac{{a}}{\:\sqrt{{a}^{\mathrm{2}} +\mathrm{8}{bc}}}+\frac{{b}}{\:\sqrt{{b}^{\mathrm{2}} +\mathrm{8}{ac}}}+\frac{{c}}{\:\sqrt{{c}^{\mathrm{2}} +\mathrm{8}{ab}}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 21/Feb/19
trying to solve...  ((a^2 +8bc)/2)≥(√(a^2 ×8bc))  a^2 +8bc≥(√(32abc×a))  (√(a^2 +8bc)) ≥(32abc×a)^(1/4)   (1/( (√(a^2 +8bc))))≤(1/((32abc×a)^(1/4) ))  (a/( (√(a^2 +8bc))))≤(a/((32abc×a)^(1/4) ))  (a/( (√(a^2 +8bc))))+(b/( (√(b^2 +8ac))))+(c/( (√(c^2 +8ab))))≤((a^(3/4) +b^(3/4) +c^(3/4) )/((32abc)^(1/4) ))  ((a^(3/4) +b^(3/4) +c^(3/4) )/3)×(1/((32abc)^(1/4) ))≥(1/((32abc)^(1/4) ))×(a^(3/4) b^(3/4) c^(3/4) )^(1/3)   =(1/((32)^(1/4) ))
$${trying}\:{to}\:{solve}… \\ $$$$\frac{{a}^{\mathrm{2}} +\mathrm{8}{bc}}{\mathrm{2}}\geqslant\sqrt{{a}^{\mathrm{2}} ×\mathrm{8}{bc}} \\ $$$${a}^{\mathrm{2}} +\mathrm{8}{bc}\geqslant\sqrt{\mathrm{32}{abc}×{a}} \\ $$$$\sqrt{{a}^{\mathrm{2}} +\mathrm{8}{bc}}\:\geqslant\left(\mathrm{32}{abc}×{a}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +\mathrm{8}{bc}}}\leqslant\frac{\mathrm{1}}{\left(\mathrm{32}{abc}×{a}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} } \\ $$$$\frac{{a}}{\:\sqrt{{a}^{\mathrm{2}} +\mathrm{8}{bc}}}\leqslant\frac{{a}}{\left(\mathrm{32}{abc}×{a}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} } \\ $$$$\frac{{a}}{\:\sqrt{{a}^{\mathrm{2}} +\mathrm{8}{bc}}}+\frac{{b}}{\:\sqrt{{b}^{\mathrm{2}} +\mathrm{8}{ac}}}+\frac{{c}}{\:\sqrt{{c}^{\mathrm{2}} +\mathrm{8}{ab}}}\leqslant\frac{{a}^{\frac{\mathrm{3}}{\mathrm{4}}} +{b}^{\frac{\mathrm{3}}{\mathrm{4}}} +{c}^{\frac{\mathrm{3}}{\mathrm{4}}} }{\left(\mathrm{32}{abc}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} } \\ $$$$\frac{{a}^{\frac{\mathrm{3}}{\mathrm{4}}} +{b}^{\frac{\mathrm{3}}{\mathrm{4}}} +{c}^{\frac{\mathrm{3}}{\mathrm{4}}} }{\mathrm{3}}×\frac{\mathrm{1}}{\left(\mathrm{32}{abc}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} }\geqslant\frac{\mathrm{1}}{\left(\mathrm{32}{abc}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} }×\left({a}^{\frac{\mathrm{3}}{\mathrm{4}}} {b}^{\frac{\mathrm{3}}{\mathrm{4}}} {c}^{\frac{\mathrm{3}}{\mathrm{4}}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{32}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} } \\ $$$$ \\ $$$$ \\ $$

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