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Question Number 144951 by loveineq last updated on 30/Jun/21

Let a,b > 0 and a+b+1 = 3ab. Prove that (1) (a^2 /(a+1))+(b^2 /(b+1)) ≥ (a/(a^2 +1))+(b/(b^2 +1)) (2) (a^2 /(b+1))+(b^2 /(a+1)) ≥ (a/(b^2 +1))+(b/(a^2 +1))

Leta,b>0anda+b+1=3ab.Provethat(1)a2a+1+b2b+1⩾aa2+1+bb2+1(2)a2b+1+b2a+1⩾ab2+1+ba2+1

Commented by justtry last updated on 01/Jul/21

(1) remember AM≥GM (1+a^2 )≥ 2a (1+a^2 )≥(1+a)^2 −(1+a^2 ) (1+a^2 )+(1+a^2 )≥(1+a)^2 −(1+a^2 )+(1+a^2 ) 2(a^2 +1)≥(1+a)^2 (1/((a+1)^2 ))≥(1/(2(a^2 +1))) .....(i) multply (i) with a^2 and (a+1) (a^2 /((a+1)^2 ))≥(a^2 /(2(a^2 +1)))≥(a/(2(a^2 +1))) ((a^2 (a+1))/((a+1)^2 ))≥((a(a+1))/(2(a^2 +1))) (a^2 /((a+1)))≥((a^2 +a)/(2(a^2 +1)))≥((a(√a))/((a^2 +1)))≥(a/((a^2 +1))) →AM≥GM ((a^2 +a)/2)≥a(√a) →a+b+1=3ab, choose a=1 ⇒ a(√a)≥a so (a^2 /(a+1))≥(a/(a^2 +1)) ...(ii) similary for b (b^2 /(b+1))≥(b/(b^2 +1)) ...(iii) from (ii) &(iii) (a^2 /(a+1)) + (b^2 /(b+1)) ≥ (a/(a^2 +1)) + (b/(b^2 +1))

(1)rememberAM⩾GM(1+a2)⩾2a(1+a2)⩾(1+a)2−(1+a2)(1+a2)+(1+a2)⩾(1+a)2−(1+a2)+(1+a2)2(a2+1)⩾(1+a)21(a+1)2⩾12(a2+1).....(i)multply(i)witha2and(a+1)a2(a+1)2⩾a22(a2+1)⩾a2(a2+1)a2(a+1)(a+1)2⩾a(a+1)2(a2+1)a2(a+1)⩾a2+a2(a2+1)⩾aa(a2+1)⩾a(a2+1)→AM⩾GMa2+a2⩾aa→a+b+1=3ab,choosea=1⇒aa⩾asoa2a+1⩾aa2+1...(ii)similaryforbb2b+1⩾bb2+1...(iii)from(ii)&(iii)a2a+1+b2b+1⩾aa2+1+bb2+1

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