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Question Number 80039 by jagoll last updated on 30/Jan/20
prove that (1+x)(1+(1/x))≥4
provethat(1+x)(1+1x)⩾4
Commented by mr W last updated on 30/Jan/20
only for x∈R^+ 1+x≥2(√x) 1+(1/x)≥2(1/( (√x))) (1+x)(1+(1/x))≥2(√x)×2(1/( (√x)))=4 or (1+x)(1+(1/x))=2+x+(1/x)≥2+2=4 for x∈R^− (1+x)(1+(1/x))=2+x+(1/x)=2−(−x+(1/(−x)))≤2−2=0
onlyforx∈R+1+x⩾2x1+1x⩾21x(1+x)(1+1x)⩾2x×21x=4or(1+x)(1+1x)=2+x+1x⩾2+2=4forx∈R−(1+x)(1+1x)=2+x+1x=2−(−x+1−x)⩽2−2=0
Commented by Tony Lin last updated on 30/Jan/20
[1^2 +((√x))^2 ][1^2 +((1/( (√x))))^2 ]≥(1+1)^2 =4
[12+(x)2][12+(1x)2]⩾(1+1)2=4
Commented by MJS last updated on 30/Jan/20
y=(1+x)(1+(1/x)) defined for x∈R\{0} (1+x)(1+(1/x))<4 x+(1/x)<2 (1) x>0 x^2 −2x+1<0 (x−1)^2 <0 wrong ⇒ (1+x)(1+(1/x))≥4 (2) x<0 x^2 −2x+1>0 (x−1)^2 >0 true ∀x<0 ⇒ (1+x)(1+(1/x))<4 { (((1+x)(1+(1/x))≥4; x>0)),(((1+x)(1+(1/x))<4; x<0)) :}
y=(1+x)(1+1x)definedforx∈R∖{0}(1+x)(1+1x)<4x+1x<2(1)x>0x2−2x+1<0(x−1)2<0wrong⇒(1+x)(1+1x)⩾4(2)x<0x2−2x+1>0(x−1)2>0true∀x<0⇒(1+x)(1+1x)<4{(1+x)(1+1x)⩾4;x>0(1+x)(1+1x)<4;x<0
Commented by mr W last updated on 30/Jan/20
typo sir? (1+x)(1+(1/x))<4; x<0 should be: (1+x)(1+(1/x))≤0; x<0
typosir?(1+x)(1+1x)<4;x<0shouldbe:(1+x)(1+1x)⩽0;x<0
Commented by MJS last updated on 30/Jan/20
You are right but with my method I only examined if it′s <4.
YouarerightbutwithmymethodIonlyexaminedifit′s<4.
Commented by jagoll last updated on 31/Jan/20
thank you
thankyou
Answered by $@ty@m123 last updated on 31/Jan/20
Let y=(1+x)(1+(1/x)) ⇒y=2+x+(1/x) (dy/dx)=1−(1/x^2 ) For max. or min. (dy/dx)=0 1−(1/x^2 )=0 ⇒x=±1 y∣_(x=1) =4, y∣_(x=−1) =0 (d^2 y/dx^2 )=(1/x^3 ) [(d^2 y/dx^2 )]_(x=1) =1>0 ∴ y is mininmum at x=1 and min.(y)=4
Lety=(1+x)(1+1x)⇒y=2+x+1xdydx=1−1x2Formax.ormin.dydx=01−1x2=0⇒x=±1y∣x=1=4,y∣x=−1=0d2ydx2=1x3[d2ydx2]x=1=1>0∴yismininmumatx=1andmin.(y)=4

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