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Question-151905

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Question Number 151905 by mathdanisur last updated on 24/Aug/21
Answered by mindispower last updated on 24/Aug/21
8x^x y^y z^z 2^(x+y+z) =2(2x)^x .2(2y)^y .(2z)^z_ 2(2t)^t ≥(t+1)^(t+1) ,t>0 ⇔2^(t+1) ≥t(1+(1/t))^(t+1) ⇔(t+1)ln(2)≥ln(t)+(t+1)ln(1+(1/t)) f(t)=(t+1)ln(2)−ln(t)−(t+1)ln(1+(1/t)) f′(t)=ln(2)−(1/t)−ln(1+(1/t))+(1/t)=ln(((2t)/(t+1))) f′(t)≥0,t∈[1,+∞[^� ,f′(t)<0,t∈]0,1[ ⇒f(t)≥f(1)=0 ⇒∀t>0,2.(2t)^(t+1) ≥(t+1)^(t+1) ⇒2.(2x)^(x+1) .2(2y)^(y+1) .2(2z)^(z+1) ≥(x+1)^(x+1) (y+1)^(y+1) (z+1)^(z+1) ⇔8x^x .y^y .z^z .2^(x+y+z) ≥(x+1)^(x+1) (y+1)^(y+1) (z+1)^(z+1)
8xxyyzz2x+y+z=2(2x)x.2(2y)y.(2z)z2(2t)t⩾(t+1)t+1,t>0⇔2t+1⩾t(1+1t)t+1⇔(t+1)ln(2)⩾ln(t)+(t+1)ln(1+1t)f(t)=(t+1)ln(2)−ln(t)−(t+1)ln(1+1t)f′(t)=ln(2)−1t−ln(1+1t)+1t=ln(2tt+1)Extra \left or missing \right⇒f(t)⩾f(1)=0⇒∀t>0,2.(2t)t+1⩾(t+1)t+1⇒2.(2x)x+1.2(2y)y+1.2(2z)z+1⩾(x+1)x+1(y+1)y+1(z+1)z+1⇔8xx.yy.zz.2x+y+z⩾(x+1)x+1(y+1)y+1(z+1)z+1
Commented by mathdanisur last updated on 24/Aug/21
Thank you Ser
ThankyouSer
Commented by mindispower last updated on 28/Aug/21
plrasur
plrasur

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