Question Number 139190 by mathdanisur last updated on 23/Apr/21
$${a};{b}\in\mathbb{R}\:,\:\frac{\mid{ax}+{b}\mid}{\mathrm{1}+{x}^{\mathrm{2}} }\:\leqslant\:\mathrm{1}\:,\:\forall{x}\in\mathbb{R} \\ $$$${prove}:\:\:\mid{a}\mid\leqslant\mathrm{2}\:;\:\mid{b}\mid\leqslant\mathrm{1} \\ $$
Answered by mitica last updated on 24/Apr/21
$${x}=\mathrm{0}\Rightarrow\mid{b}\mid\leqslant\mathrm{1} \\ $$$${x}=\mathrm{1}\Rightarrow\mid{a}+{b}\mid\leqslant\mathrm{2} \\ $$$${x}=−\mathrm{1}\Rightarrow\mid{a}−{b}\mid\leqslant\mathrm{2} \\ $$$$\mid{a}\mid=\mid\frac{\mathrm{2}{a}}{\mathrm{2}}\mid=\frac{\mid{a}+{b}+{a}−{b}\mid}{\mathrm{2}}\leqslant\frac{\mid{a}+{b}\mid+\mid{a}−{b}\mid}{\mathrm{2}}\leqslant\frac{\mathrm{2}+\mathrm{2}}{\mathrm{2}}=\mathrm{2} \\ $$
Commented by mathdanisur last updated on 24/Apr/21
$${thank}\:{you}\:{sir},\:{but}\:\mid{b}\mid\leqslant\mathrm{1}?\:{please} \\ $$
Commented by mitica last updated on 24/Apr/21
$${x}=\mathrm{0}\Rightarrow\frac{\mid{a}\centerdot\mathrm{0}+{b}\mid}{\mathrm{0}^{\mathrm{2}} +\mathrm{1}}\leqslant\mathrm{1}\Rightarrow\frac{\mid{b}\mid}{\mathrm{1}}\leqslant\mathrm{1}\Rightarrow\mid{b}\mid\leqslant\mathrm{1} \\ $$
Answered by mr W last updated on 24/Apr/21
Commented by mr W last updated on 24/Apr/21
$${let}'{s}\:{generally}\:{look}\:{at}\:{the}\:{function} \\ $$$${y}={x}^{\mathrm{2}} +{bx}+{c} \\ $$$${there}\:{are}\:{three}\:{cases}. \\ $$$${case}\:\mathrm{1}:\:{y}>\mathrm{0}\:{for}\:{x}\in{R} \\ $$$${x}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\:{has}\:{no}\:{real}\:{root},\:{i}.{e}. \\ $$$$\Delta={b}^{\mathrm{2}} −\mathrm{4}{c}<\mathrm{0} \\ $$$${case}\:\mathrm{2}:\:{y}\geqslant\mathrm{0}\:{for}\:{x}\in{R} \\ $$$${x}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\:{has}\:{one}\:{double}\:{real}\:{root},\:{i}.{e}. \\ $$$$\Delta={b}^{\mathrm{2}} −\mathrm{4}{c}=\mathrm{0} \\ $$$${case}\:\mathrm{3}: \\ $$$${x}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\:{has}\:{two}\:{real}\:{rootx},\:{i}.{e}. \\ $$$$\Delta={b}^{\mathrm{2}} −\mathrm{4}{c}>\mathrm{0} \\ $$$${y}>\mathrm{0}\:{if}\:{x}<{x}_{\mathrm{1}} \:{or}\:{x}>{x}_{\mathrm{2}} \\ $$$${y}<\mathrm{0}\:{if}\:{x}_{\mathrm{1}} <{x}<{x}_{\mathrm{2}} \\ $$$${y}=\mathrm{0}\:{if}\:{x}={x}_{\mathrm{1}} \:{or}\:{x}_{\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow{such}\:{that}\:{x}^{\mathrm{2}} +{bx}+{c}\geqslant\mathrm{0}\:{for}\:{x}\in{R}, \\ $$$$\Delta={b}^{\mathrm{2}} −\mathrm{4}{c}\leqslant\mathrm{0} \\ $$
Commented by mr W last updated on 24/Apr/21
$$\frac{\mid{ax}+{b}\mid}{\mathrm{1}+{x}^{\mathrm{2}} }\leqslant\mathrm{1}\:{for}\:{x}\in{R} \\ $$$$\Rightarrow\:−\mathrm{1}\leqslant\frac{{ax}+{b}}{\mathrm{1}+{x}^{\mathrm{2}} }\leqslant\mathrm{1} \\ $$$$\Rightarrow\:−\mathrm{1}−{x}^{\mathrm{2}} \leqslant{ax}+{b}\leqslant\mathrm{1}+{x}^{\mathrm{2}} \\ $$$$ \\ $$$$\left.\mathrm{1}\right)\:{ax}+{b}\leqslant\mathrm{1}+{x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −{ax}+\mathrm{1}−{b}\geqslant\mathrm{0} \\ $$$$\Delta={a}^{\mathrm{2}} +\mathrm{4}\left({b}−\mathrm{1}\right)\leqslant\mathrm{0}\:\:\:…\left({i}\right) \\ $$$$\mathrm{4}\left({b}−\mathrm{1}\right)\leqslant\mathrm{0} \\ $$$$\Rightarrow{b}\leqslant\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:−\mathrm{1}−{x}^{\mathrm{2}} \leqslant{ax}+{b} \\ $$$${x}^{\mathrm{2}} +{ax}+{b}+\mathrm{1}\geqslant\mathrm{0}\:\:\: \\ $$$$\Delta={a}^{\mathrm{2}} −\mathrm{4}\left({b}+\mathrm{1}\right)\leqslant\mathrm{0}\:\:\:…\left({ii}\right) \\ $$$$\mathrm{4}\left({b}+\mathrm{1}\right)\geqslant{a}^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\Rightarrow{b}\geqslant−\mathrm{1} \\ $$$$\Rightarrow−\mathrm{1}\leqslant{b}\leqslant\mathrm{1} \\ $$$$\Rightarrow\mid{b}\mid\leqslant\mathrm{1} \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$$\mathrm{2}{a}^{\mathrm{2}} −\mathrm{4}−\mathrm{4}\leqslant\mathrm{0} \\ $$$${a}^{\mathrm{2}} \leqslant\mathrm{4} \\ $$$$\Rightarrow\mid{a}\mid\leqslant\mathrm{2} \\ $$
Commented by mathdanisur last updated on 24/Apr/21
$${Perfect}\:{Sir}\:{thanks}.. \\ $$$${How}\:{can}\:{this}\:{be}\:{Sir}\:{pliase}.. \\ $$$${if}:\:{a};{b}\in\mathbb{R},\:\mid{ax}^{\mathrm{3}} +{bx}\mid\leqslant\mathrm{1},\:\forall\mid{x}\mid\leqslant\mathrm{1} \\ $$$${prov}:\:\mid{bx}^{\mathrm{3}} +{ax}\mid\leqslant\mathrm{1},\:\mid\mathrm{3}{ax}^{\mathrm{2}} +{b}\mid\leqslant\mathrm{9},\:\forall\mid{x}\mid\leqslant\mathrm{1} \\ $$
Commented by mr W last updated on 24/Apr/21
$${please}\:{open}\:{a}\:{new}\:{thread}\:{for}\:{this}\: \\ $$$${new}\:{question}! \\ $$