a;b∈R , ((∣ax+b∣)/(1+x^2 )) ≤ 1 , ∀x∈R prove: ∣a∣≤2 ; ∣b∣≤1


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Question Number 139190 by mathdanisur last updated on 23/Apr/21

$a; b \in R, \frac{∣ a x + b ∣}{1 + x^{2}} ⩽ 1, \forall x \in R$ $p r o v e : ∣ a ∣⩽ 2; ∣ b ∣⩽ 1$
	Answered by mitica last updated on 24/Apr/21

	$x = 0 \Rightarrow∣ b ∣⩽ 1$ $x = 1 \Rightarrow∣ a + b ∣⩽ 2$ $x = - 1 \Rightarrow∣ a - b ∣⩽ 2$ $∣ a ∣=∣ \frac{2 a}{2} ∣= \frac{∣ a + b + a - b ∣}{2} ⩽ \frac{∣ a + b ∣ + ∣ a - b ∣}{2} ⩽ \frac{2 + 2}{2} = 2$
	Commented by mathdanisur last updated on 24/Apr/21

	$t h a n k y o u s i r, b u t ∣ b ∣⩽ 1 ? p l e a s e$
	Commented by mitica last updated on 24/Apr/21

	$x = 0 \Rightarrow \frac{∣ a \cdot 0 + b ∣}{0^{2} + 1} ⩽ 1 \Rightarrow \frac{∣ b ∣}{1} ⩽ 1 \Rightarrow∣ b ∣⩽ 1$
	Answered by mr W last updated on 24/Apr/21

	Commented by mr W last updated on 24/Apr/21

	${l e t}^{'} s g e n e r a l l y l o o k a t t h e f u n c t i o n$ $y = x^{2} + b x + c$ $t h e r e a r e t h r e e c a s e s .$ $c a s e 1 : y > 0 f o r x \in R$ $x^{2} + b x + c = 0 h a s n o r e a l r o o t, i . e .$ $Δ = b^{2} - 4 c < 0$ $c a s e 2 : y ⩾ 0 f o r x \in R$ $x^{2} + b x + c = 0 h a s o n e d o u b l e r e a l r o o t, i . e .$ $Δ = b^{2} - 4 c = 0$ $c a s e 3 :$ $x^{2} + b x + c = 0 h a s t w o r e a l r o o t x, i . e .$ $Δ = b^{2} - 4 c > 0$ $y > 0 i f x < x_{1} o r x > x_{2}$ $y < 0 i f x_{1} < x < x_{2}$ $y = 0 i f x = x_{1} o r x_{2}$ $\Rightarrow s u c h t h a t x^{2} + b x + c ⩾ 0 f o r x \in R,$ $Δ = b^{2} - 4 c ⩽ 0$
	Commented by mr W last updated on 24/Apr/21

	$\frac{∣ a x + b ∣}{1 + x^{2}} ⩽ 1 f o r x \in R$ $\Rightarrow - 1 ⩽ \frac{a x + b}{1 + x^{2}} ⩽ 1$ $\Rightarrow - 1 - x^{2} ⩽ a x + b ⩽ 1 + x^{2}$ $1) a x + b ⩽ 1 + x^{2}$ $x^{2} - a x + 1 - b ⩾ 0$ $Δ = a^{2} + 4 (b - 1) ⩽ 0 . . . (i)$ $4 (b - 1) ⩽ 0$ $\Rightarrow b ⩽ 1$ $2) - 1 - x^{2} ⩽ a x + b$ $x^{2} + a x + b + 1 ⩾ 0$ $Δ = a^{2} - 4 (b + 1) ⩽ 0 . . . (i i)$ $4 (b + 1) ⩾ a^{2} ⩾ 0$ $\Rightarrow b ⩾ - 1$ $\Rightarrow - 1 ⩽ b ⩽ 1$ $\Rightarrow∣ b ∣⩽ 1$ $(i) + (i i) :$ $2 a^{2} - 4 - 4 ⩽ 0$ $a^{2} ⩽ 4$ $\Rightarrow∣ a ∣⩽ 2$
	Commented by mathdanisur last updated on 24/Apr/21

	$P e r f e c t S i r t h a n k s . .$ $H o w c a n t h i s b e S i r p l i a s e . .$ $i f : a; b \in R, ∣ {a x}^{3} + b x ∣⩽ 1, \forall ∣ x ∣⩽ 1$ $p r o v : ∣ {b x}^{3} + a x ∣⩽ 1, ∣ 3 {a x}^{2} + b ∣⩽ 9, \forall ∣ x ∣⩽ 1$
	Commented by mr W last updated on 24/Apr/21

	$p l e a s e o p e n a n e w t h r e a d f o r t h i s$ $n e w q u e s t i o n!$
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