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Question Number 151351 by mathdanisur last updated on 20/Aug/21

Answered by dumitrel last updated on 20/Aug/21

λ(√((a^2 +b^2 )/2))+((2ab)/(a+b))≥((λ+1)/2)(a+b) if a=b true suppose a<b ;(a/b)=t∈(0;1) ⇔λ(√((t^2 +1)/2))+((2t)/(t+1))≥(((λ+1)(t+1))/2) ⇔λ((((√(2t^2 +2))−t−1)/2))≥((t+1)/2)−((2t)/(t+1))⇔ λ∙(((λ−1)^2 )/( (√(2t^2 +2))+t+1))≥(((t−1)^2 )/(t+1))⇔λ≥(((√(2(t^2 +1)))+t+1)/(t+1)) true (√2)>((√(2(t^2 +1)))/(t+1)) λ≥(√2)+1>((√(2(t^2 +1)))/(t+1))+((t+1)/(t+1))=(((√(2(t^2 +1)))+t+1)/(t+1))

λa2+b22+2aba+b⩾λ+12(a+b)ifa=btruesupposea<b;ab=t∈(0;1)⇔λt2+12+2tt+1⩾(λ+1)(t+1)2⇔λ(2t2+2−t−12)⩾t+12−2tt+1⇔λ⋅(λ−1)22t2+2+t+1⩾(t−1)2t+1⇔λ⩾2(t2+1)+t+1t+1true2>2(t2+1)t+1λ⩾2+1>2(t2+1)t+1+t+1t+1=2(t2+1)+t+1t+1

Commented by mathdanisur last updated on 20/Aug/21

Nice Ser, Thank You

NiceSer,ThankYou

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