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Question Number 142906 by liberty last updated on 07/Jun/21
If abc=1 and a,b,c>0 prove that (a/(b^2 (c+1)))+(b/(c^2 (a+1)))+(c/(a^2 (b+1))) ≥ (3/2)
$${If}\:{abc}=\mathrm{1}\:{and}\:{a},{b},{c}>\mathrm{0}\:{prove} \\ $$$${that}\:\frac{{a}}{{b}^{\mathrm{2}} \left({c}+\mathrm{1}\right)}+\frac{{b}}{{c}^{\mathrm{2}} \left({a}+\mathrm{1}\right)}+\frac{{c}}{{a}^{\mathrm{2}} \left({b}+\mathrm{1}\right)}\:\geqslant\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Answered by Snail last updated on 07/Jun/21
Let us recall Titu′s Lemma Σ(x_i ^2 /y_i )≥ (((Σx_i )^2 )/(Σy_i )) Denote give ineauality by E As abc=1 Multiplying the inequality both side by a^2 b^2 c^2 and using its valud 1 we ca transform it and get ... ((a^3 c^2 )/(c+1))+((b^3 a^2 )/(a+1))+((c^3 b^2 )/(b+1))≥(3/2) now let x_1 =a^2 c^2 x_2 =b^2 a^2 x_3 =c^2 b^2 y_1 =((c+1)/a) y_2 =((a+1)/b) y_3 =((b+1)/c) E ≥(((a^2 c^2 +b^2 c^2 +c^2 a^2 )^2 )/(((c+1)/a)+((b+1)/c)+((a+1)/b))) =Z(Let) Now usi ng the fact that (a^2 c^2 +b^2 c^2 +a^2 b^2 )^2 ≥9 and by doing LCM of denominator we can get that Z =(3/(2 )) So E≥Z equality holds when all (x_i /y_i ) are same Proved....
$${Let}\:{us}\:{recall}\:{Titu}'{s}\:{Lemma} \\ $$$$\Sigma\frac{{x}_{{i}} ^{\mathrm{2}} }{{y}_{{i}} }\geqslant\:\frac{\left(\Sigma{x}_{{i}} \right)^{\mathrm{2}} }{\Sigma{y}_{{i}} } \\ $$$${Denote}\:{give}\:\:{ineauality}\:{by}\:{E} \\ $$$${As}\:{abc}=\mathrm{1}\:\:{Multiplying}\:{the}\:{inequality}\:{both}\:{side} \\ $$$${by}\:{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} \:{and}\:{using}\:{its}\:{valud}\:\mathrm{1}\:{we}\:{ca}\:{transform} \\ $$$${it}\:{and}\:{get}\:… \\ $$$$\frac{{a}^{\mathrm{3}} {c}^{\mathrm{2}} }{{c}+\mathrm{1}}+\frac{{b}^{\mathrm{3}} {a}^{\mathrm{2}} }{{a}+\mathrm{1}}+\frac{{c}^{\mathrm{3}} {b}^{\mathrm{2}} }{{b}+\mathrm{1}}\geqslant\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${now}\:{let}\:{x}_{\mathrm{1}} ={a}^{\mathrm{2}} {c}^{\mathrm{2}} \:{x}_{\mathrm{2}} ={b}^{\mathrm{2}} {a}^{\mathrm{2}} \:{x}_{\mathrm{3}} ={c}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$${y}_{\mathrm{1}} =\frac{{c}+\mathrm{1}}{{a}}\:\:\:{y}_{\mathrm{2}} =\frac{{a}+\mathrm{1}}{{b}}\:\:{y}_{\mathrm{3}} =\frac{{b}+\mathrm{1}}{{c}} \\ $$$$ \\ $$$${E}\:\geqslant\frac{\left({a}^{\mathrm{2}} {c}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} \right)^{\mathrm{2}} }{\frac{{c}+\mathrm{1}}{{a}}+\frac{{b}+\mathrm{1}}{{c}}+\frac{{a}+\mathrm{1}}{{b}}}\:\:={Z}\left({Let}\right) \\ $$$${Now}\:{usi}\:{ng}\:{the}\:{fact}\:{that}\:\:\:\:\left({a}^{\mathrm{2}} {c}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right)^{\mathrm{2}} \geqslant\mathrm{9} \\ $$$${and}\:{by}\:{doing}\:{LCM}\:{of}\:{denominator}\:{we}\:{can}\:{get}\:{that}\: \\ $$$${Z}\:=\frac{\mathrm{3}}{\mathrm{2}\:} \\ $$$${So}\:{E}\geqslant{Z}\:{equality}\:{holds}\:{when}\:{all}\:\:\frac{{x}_{{i}} }{{y}_{{i}} }\:{are}\:{same}\: \\ $$$${Proved}…. \\ $$$$ \\ $$$$ \\ $$

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