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if-x-gt-1-q-2-then-1-x-q-1-qx-q-1-x-2-

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Question Number 147119 by mathdanisur last updated on 18/Jul/21
if x>−1 , q≥2 then: (1+x)^q ≥ 1+qx+(q−1)x^2
ifx>−1,q⩾2then:(1+x)q⩾1+qx+(q−1)x2
Answered by mindispower last updated on 18/Jul/21
1+qx+(q−1)x^2 =(q−1)(x+1)(x+(1/(q−1))) ⇔(((1+x)^(q−1) )/(q−1))≥x+(1/(q−1)) ⇔(1+x)^(q−1) ≥(q−1)x+1 f(x)=(1+x)^(q−1) −((q−1)x+1) f′(x)=(q−1)((1+x)^(q−2) −1) f′(x)>0,x∈[0,∞[ f′(x)<0,x∈]−1,0[ ⇒∀x∈]−1,+∞[ ⇒f(x)≥f(0)=(1+0)^(q−1) −1=0 ⇒f(x)≥0⇔(1+x)^(q−1) ≥(q−1)x+1
1+qx+(q−1)x2=(q−1)(x+1)(x+1q−1)⇔(1+x)q−1q−1⩾x+1q−1⇔(1+x)q−1⩾(q−1)x+1f(x)=(1+x)q−1−((q−1)x+1)f′(x)=(q−1)((1+x)q−2−1)f′(x)>0,x∈[0,∞[f′(x)<0,x∈]−1,0[⇒∀x∈]−1,+∞[⇒f(x)⩾f(0)=(1+0)q−1−1=0⇒f(x)⩾0⇔(1+x)q−1⩾(q−1)x+1
Commented by mathdanisur last updated on 19/Jul/21
thank you Ser
thankyouSer
Answered by mathmax by abdo last updated on 19/Jul/21
⇒(1+x)^n ≥1+nx+(n−1)x^2 ⇒ p_n (x)=(1+x)^n −1−nx−(n−1)x^2 ≥0 (p_n )by recurrence on n n=2 ⇒(1+x)^2 −1−2x−x^2 =0≥0 true let suppose p_n (x)≥0 and prove p_(n+1) (x)≥0 p_(n+1) (x)=(1+x)^(n+1) −1−(n+1)x−nx^2 =(1+x)(1+x)^n −1−(n+1)x−nx^2 =(1+x){p_n (x)+1+nx+(n−1)x^2 }−1−(n+1)x−nx^2 =p_n (x)+1+nx+(n−1)x^2 +xp_n (x)+x+nx^2 +(n−1)x^3 −1−(n+1)x−nx^2 =(1+x)p_n (x)+(n−1)x^(2 ) +(n−1)x^3 ≥0 due to p_n ≥0 and n−1≥1
⇒(1+x)n⩾1+nx+(n−1)x2⇒pn(x)=(1+x)n−1−nx−(n−1)x2⩾0(pn)byrecurrenceonnn=2⇒(1+x)2−1−2x−x2=0⩾0trueletsupposepn(x)⩾0andprovepn+1(x)⩾0pn+1(x)=(1+x)n+1−1−(n+1)x−nx2=(1+x)(1+x)n−1−(n+1)x−nx2=(1+x){pn(x)+1+nx+(n−1)x2}−1−(n+1)x−nx2=pn(x)+1+nx+(n−1)x2+xpn(x)+x+nx2+(n−1)x3−1−(n+1)x−nx2=(1+x)pn(x)+(n−1)x2+(n−1)x3⩾0duetopn⩾0andn−1⩾1
Commented by mathdanisur last updated on 19/Jul/21
thank you Ser
thankyouSer

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