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Question Number 205733 by hardmath last updated on 28/Mar/24

If a,b,c∈R^+ and a+b+c=6 Prove that: ((a^2 −4)/(4a^2 −9a + 6)) + ((b^2 −4)/(4b^2 −9b + 6)) + ((c^2 −4)/(4c^2 −9c + 6)) ≤ 0

Ifa,b,c∈R+anda+b+c=6Provethat:a2−44a2−9a+6+b2−44b2−9b+6+c2−44c2−9c+6⩽0

Answered by A5T last updated on 29/Mar/24

4a^2 +16≥2(√(4×16a^2 ))=16a ⇒4a^2 +16−9a−10≥7a−10 ((a^2 −4=(a/7)(7a−10)+((10a)/7)−4)/(7a−10))=(a/7)+(((10a−28)/7)/(7a−10)) Question⇔Σ((a/7)+((10a−28)/(49a−70)))≤0 ⇔Σ((10a−28)/(49a−70))≤((−6)/7)⇔((10)/(49))×3−Σ((10a−28)/(49a−70))≥((72)/(49)) ⇔Σ(((672)/(343(7a−10))))≥((72)/(49))⇔((672)/(343))(Σ(1/((7a−10))))≥((72)/(49)) ⇔^? Σ(1/(7a−10))≥((72×343)/(49×672))=(3/4) (1/(7a−10))+(1/(7b−10))+(1/(7c−10))≥(9/(7(a+b+c)−30))=(9/(12))=(3/4) Since Σ(1/(7a−10))≥(3/4) as shown above,then original ineauality must be true[Equality when a=b=c=2]

4a2+16⩾24×16a2=16a⇒4a2+16−9a−10⩾7a−10a2−4=a7(7a−10)+10a7−47a−10=a7+10a−2877a−10Question⇔Σ(a7+10a−2849a−70)⩽0⇔Σ10a−2849a−70⩽−67⇔1049×3−Σ10a−2849a−70⩾7249⇔Σ(672343(7a−10))⩾7249⇔672343(Σ1(7a−10))⩾7249⇔?Σ17a−10⩾72×34349×672=3417a−10+17b−10+17c−10⩾97(a+b+c)−30=912=34SinceΣ17a−10⩾34asshownabove,thenoriginalineaualitymustbetrue[Equalitywhena=b=c=2]

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