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Question Number 59273 by naka3546 last updated on 07/May/19

abc = 64 a, b, c ∈ R^+ Find K that satisfy to the inequality : (((a + b) (√(ab)) + (b + c) (√(bc)) + (c + a) (√(ca)))/(√(abc))) ≥ (√a) + (√b) + (√c) + K .

$${abc}\:\:=\:\:\mathrm{64} \\ $$$${a},\:{b},\:{c}\:\:\in\:\:\mathbb{R}^{+} \\ $$$${Find}\:\:{K}\:\:{that}\:\:{satisfy}\:\:{to}\:\:{the}\:\:{inequality}\:\:: \\ $$$$\:\:\:\:\:\frac{\left({a}\:+\:{b}\right)\:\sqrt{{ab}}\:\:+\:\:\left({b}\:+\:{c}\right)\:\sqrt{{bc}}\:\:+\:\:\left({c}\:+\:{a}\right)\:\sqrt{{ca}}}{\sqrt{{abc}}}\:\:\:\geqslant\:\:\sqrt{{a}}\:\:+\:\:\sqrt{{b}}\:\:+\:\:\sqrt{{c}}\:\:+\:\:{K}\:\:. \\ $$

Answered by tanmay last updated on 07/May/19

((a+b)/2)≥(√(ab)) (a+b)≥2(√(ab)) (a+b)(√(ab)) ≥2ab so (a+b)(√(ab)) +(b+c)(√(bc)) +(c+a)(√(ac)) ≥2(ab+bc+ca) LHS (N_r /D_r )≥((2(ab+bc+ca))/(√(abc))) (N_r /D_r )≥((ab+bc+ca)/4) ((a+b+c)/3)≥(abc)^(1/3) (a+b+c)≥12 ★now ((ab+bc+ca)/3)≥(ab×bc×ca)^(1/3) ((ab+bc+ca)/3)≥(abc)^(2/3) ((ab+bc+ca)/4)≥(3/4)×(64)^(2/3) ((ab+bc+ca)/4)≥(3/4)×16 ((ab+bc+ca)/4)≥12★ (N_r /D_r )≥((ab+bc+ca)/4) (N_r /D_r )≥((ab+bc+ca)/4)≥12 RHS (√a) +(√b) +(√c) +(√k) (((√a) +(√b) +(√c) )/3)≥((√a) ×(√b) ×(√c) )^(1/3) (((√a) +(√b) +(√c) )/3)≥(abc)^(1/6) ((√a) +(√b) +(√c) )/3 ≥(64)^(1/6) ←correction after Naka sir detected (√a) +(√b) +(√c) ≥2×3=6 so (N_r /D_r )≥((ab+bc+ca)/4)≥12≥6+k so when 12=6+k k=6 when 12≥6+k 6≥k i have just tried...

$$\frac{{a}+{b}}{\mathrm{2}}\geqslant\sqrt{{ab}}\: \\ $$$$\left({a}+{b}\right)\geqslant\mathrm{2}\sqrt{{ab}}\: \\ $$$$\left({a}+{b}\right)\sqrt{{ab}}\:\geqslant\mathrm{2}{ab} \\ $$$${so}\:\left({a}+{b}\right)\sqrt{{ab}}\:+\left({b}+{c}\right)\sqrt{{bc}}\:+\left({c}+{a}\right)\sqrt{{ac}}\:\geqslant\mathrm{2}\left({ab}+{bc}+{ca}\right) \\ $$$${LHS} \\ $$$$\frac{{N}_{{r}} }{{D}_{{r}} }\geqslant\frac{\mathrm{2}\left({ab}+{bc}+{ca}\right)}{\sqrt{{abc}}} \\ $$$$\frac{{N}_{{r}} }{{D}_{{r}} }\geqslant\frac{{ab}+{bc}+{ca}}{\mathrm{4}} \\ $$$$\frac{{a}+{b}+{c}}{\mathrm{3}}\geqslant\left({abc}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\left({a}+{b}+{c}\right)\geqslant\mathrm{12} \\ $$$$\bigstar{now}\:\frac{{ab}+{bc}+{ca}}{\mathrm{3}}\geqslant\left({ab}×{bc}×{ca}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\frac{{ab}+{bc}+{ca}}{\mathrm{3}}\geqslant\left({abc}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$\frac{{ab}+{bc}+{ca}}{\mathrm{4}}\geqslant\frac{\mathrm{3}}{\mathrm{4}}×\left(\mathrm{64}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$\frac{{ab}+{bc}+{ca}}{\mathrm{4}}\geqslant\frac{\mathrm{3}}{\mathrm{4}}×\mathrm{16} \\ $$$$\frac{{ab}+{bc}+{ca}}{\mathrm{4}}\geqslant\mathrm{12}\bigstar \\ $$$$\frac{{N}_{{r}} }{{D}_{{r}} }\geqslant\frac{{ab}+{bc}+{ca}}{\mathrm{4}} \\ $$$$\frac{{N}_{{r}} }{{D}_{{r}} }\geqslant\frac{{ab}+{bc}+{ca}}{\mathrm{4}}\geqslant\mathrm{12} \\ $$$${RHS} \\ $$$$\sqrt{{a}}\:+\sqrt{{b}}\:+\sqrt{{c}}\:+\sqrt{{k}}\: \\ $$$$\frac{\sqrt{{a}}\:+\sqrt{{b}}\:+\sqrt{{c}}\:}{\mathrm{3}}\geqslant\left(\sqrt{{a}}\:×\sqrt{{b}}\:×\sqrt{{c}}\:\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\frac{\sqrt{{a}}\:+\sqrt{{b}}\:+\sqrt{{c}}\:}{\mathrm{3}}\geqslant\left({abc}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} \\ $$$$\left(\sqrt{{a}}\:+\sqrt{{b}}\:+\sqrt{{c}}\:\right)/\mathrm{3}\:\geqslant\left(\mathrm{64}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} \:\:\:\leftarrow{correction}\:{after}\:{Naka}\:{sir}\:{detected} \\ $$$$\sqrt{{a}}\:+\sqrt{{b}}\:+\sqrt{{c}}\:\geqslant\mathrm{2}×\mathrm{3}=\mathrm{6} \\ $$$${so} \\ $$$$\frac{{N}_{{r}} }{{D}_{{r}} }\geqslant\frac{{ab}+{bc}+{ca}}{\mathrm{4}}\geqslant\mathrm{12}\geqslant\mathrm{6}+{k} \\ $$$${so}\:{when}\:\mathrm{12}=\mathrm{6}+{k}\:\:\:\:\:\:{k}=\mathrm{6} \\ $$$${when}\:\mathrm{12}\geqslant\mathrm{6}+{k} \\ $$$$\mathrm{6}\geqslant{k} \\ $$$${i}\:{have}\:{just}\:{tried}... \\ $$$$ \\ $$

Commented by naka3546 last updated on 07/May/19

(√a) + (√b) + (√c) ≥ 3∙2 = 6 ? (N_r /D_r )≥((ab+bc+ca)/4)≥12≥2+k so when 12 = 6 + k k = 6 K = 6

$$\sqrt{{a}}\:\:+\:\:\sqrt{{b}}\:\:+\:\:\sqrt{{c}}\:\:\:\geqslant\:\:\mathrm{3}\centerdot\mathrm{2}\:\:=\:\:\mathrm{6}\:\:? \\ $$$$\frac{{N}_{{r}} }{{D}_{{r}} }\geqslant\frac{{ab}+{bc}+{ca}}{\mathrm{4}}\geqslant\mathrm{12}\geqslant\mathrm{2}+{k}\:\:\: \\ $$$${so}\:{when}\:\mathrm{12}\:=\:\mathrm{6}\:+\:{k}\:\:\:\:\:\:{k}\:=\:\mathrm{6}\:\:\: \\ $$$${K}\:\:=\:\:\mathrm{6} \\ $$

Commented by tanmay last updated on 07/May/19

thank you sir for your kind perusal and detection of error...

$${thank}\:{you}\:{sir}\:{for}\:{your}\:{kind}\:{perusal}\:{and}\:{detection} \\ $$$${of}\:{error}... \\ $$

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