if x>−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

$i f x > - 1, q ⩾ 2 t h e n :$ ${(1 + x)}^{q} ⩾ 1 + q x + (q - 1) x^{2}$
	Answered by mindispower last updated on 18/Jul/21

	$1 + q x + (q - 1) x^{2} = (q - 1) (x + 1) (x + \frac{1}{q - 1})$ $\Leftrightarrow \frac{{(1 + x)}^{q - 1}}{q - 1} ⩾ x + \frac{1}{q - 1}$ $\Leftrightarrow {(1 + x)}^{q - 1} ⩾ (q - 1) x + 1$ $f (x) = {(1 + x)}^{q - 1} - ((q - 1) x + 1)$ $f^{'} (x) = (q - 1) ({(1 + x)}^{q - 2} - 1)$ $f^{'} (x) > 0, x \in [0, \infty [$ $f^{'} (x) < 0, x \in] - 1, 0 [$ $\Rightarrow \forall x \in] - 1, + \infty [$ $\Rightarrow f (x) ⩾ f (0) = {(1 + 0)}^{q - 1} - 1 = 0$ $\Rightarrow f (x) ⩾ 0 \Leftrightarrow {(1 + x)}^{q - 1} ⩾ (q - 1) x + 1$
	Commented bymathdanisur last updated on 19/Jul/21

	$t h a n k y o u S e r$
	Answered by mathmax by abdo last updated on 19/Jul/21
	$⇒(1+x)^n ≥1+nx+(n−1)x^2 ⇒ p_n (x)=(1+x)^n −1−nx−(n−1)x^2 ≥0 (p_n )by recurrence on n n=2 ⇒(1+x)^2 −1−2x−x^2 =0≥0 true let suppose p_n (x)≥0 and prove p_(n+1) (x)≥0 p_(n+1) (x)=(1+x)^(n+1) −1−(n+1)x−nx^2 =(1+x)(1+x)^n −1−(n+1)x−nx^2 =(1+x){p_n (x)+1+nx+(n−1)x^2 }−1−(n+1)x−nx^2 =p_n (x)+1+nx+(n−1)x^2 +xp_n (x)+x+nx^2 +(n−1)x^3 −1−(n+1)x−nx^2 =(1+x)p_n (x)+(n−1)x^(2 ) +(n−1)x^3 ≥0 due to p_n ≥0 and n−1≥1$
	$\Rightarrow {(1 + x)}^{n} ⩾ 1 + nx + (n - 1) x^{2} \Rightarrow$ $p_{n} (x) = {(1 + x)}^{n} - 1 - nx - (n - 1) x^{2} ⩾ 0 (p_{n}) by recurrence on n$ $n = 2 \Rightarrow {(1 + x)}^{2} - 1 - 2 x - x^{2} = 0 ⩾ 0 true$ $let suppose p_{n} (x) ⩾ 0 and prove p_{n + 1} (x) ⩾ 0$ $p_{n + 1} (x) = {(1 + x)}^{n + 1} - 1 - (n + 1) x - {nx}^{2}$ $= (1 + x) {(1 + x)}^{n} - 1 - (n + 1) x - {nx}^{2}$ $= (1 + x) {p_{n} (x) + 1 + nx + (n - 1) x^{2}} - 1 - (n + 1) x - {nx}^{2}$ $= p_{n} (x) + 1 + nx + (n - 1) x^{2} + {xp}_{n} (x) + x + {nx}^{2} + (n - 1) x^{3} - 1 - (n + 1) x - {nx}^{2}$ $= (1 + x) p_{n} (x) + (n - 1) x^{2} + (n - 1) x^{3} ⩾ 0 due to p_{n} ⩾ 0 and n - 1 ⩾ 1$
	Commented bymathdanisur last updated on 19/Jul/21

	$t h a n k y o u S e r$
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