Solve the given inequality below in the body.Rational/Quadratic InequalityProof that $left(sum^n_k=1x_kright)left(sum^n_k=1y_kright)geq n^2$Is $int_x^inftye^-fract^22 < frac1xe^-fracx^22$?Is the inequality solution below legal?Young's inequality or something similarTypical Absolute value inequalityWhich is the proper way to solve the inequality problems?Solve the inequality $sin x > ln x$
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Solve the given inequality below in the body.
Rational/Quadratic InequalityProof that $left(sum^n_k=1x_kright)left(sum^n_k=1y_kright)geq n^2$Is $int_x^inftye^-fract^22 < frac1xe^-fracx^22$?Is the inequality solution below legal?Young's inequality or something similarTypical Absolute value inequalityWhich is the proper way to solve the inequality problems?Solve the inequality $sin x > ln x$
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
$$frac1x le frac 12$$
Let’s consider $|x|=y$
So $$frac1y-3-frac 12 le 0$$
$$frac2-y+3y-3 le 0$$
$$fracy-5y-3 ge 0$$
$$y in (-infty , 3)cup [5, infty)$$
Now this is where the problem starts. I cannot figure out on how to break the modulus function, so that’s the part where I need help in. Thanks a lot!
inequality
asked 9 hours ago
Aditya Aditya
39910 bronze badges
$endgroup$
add a comment |
$begingroup$
$$frac1x le frac 12$$
Let’s consider $|x|=y$
So $$frac1y-3-frac 12 le 0$$
$$frac2-y+3y-3 le 0$$
$$fracy-5y-3 ge 0$$
$$y in (-infty , 3)cup [5, infty)$$
Now this is where the problem starts. I cannot figure out on how to break the modulus function, so that’s the part where I need help in. Thanks a lot!
inequality
asked 9 hours ago
Aditya Aditya
39910 bronze badges
$endgroup$
add a comment |
$begingroup$
$$frac1x le frac 12$$
Let’s consider $|x|=y$
So $$frac1y-3-frac 12 le 0$$
$$frac2-y+3y-3 le 0$$
$$fracy-5y-3 ge 0$$
$$y in (-infty , 3)cup [5, infty)$$
Now this is where the problem starts. I cannot figure out on how to break the modulus function, so that’s the part where I need help in. Thanks a lot!
inequality
asked 9 hours ago
Aditya Aditya
39910 bronze badges
$endgroup$
$$frac1x le frac 12$$
Let’s consider $|x|=y$
So $$frac1y-3-frac 12 le 0$$
$$frac2-y+3y-3 le 0$$
$$fracy-5y-3 ge 0$$
$$y in (-infty , 3)cup [5, infty)$$
Now this is where the problem starts. I cannot figure out on how to break the modulus function, so that’s the part where I need help in. Thanks a lot!
inequality
inequality
asked 9 hours ago
Aditya Aditya
39910 bronze badges
asked 9 hours ago
Aditya Aditya
39910 bronze badges
asked 9 hours ago
Aditya Aditya
39910 bronze badges
asked 9 hours ago
Aditya Aditya
39910 bronze badges
asked 9 hours ago
Aditya Aditya
39910 bronze badges
39910 bronze badges
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Anyway, in your case, $yge 0$, so you should transform all this to
$$|x|<3;textit or ; |x|ge 5iff -3<x<3;textit or ; xge 5;textit or ; xle -5$$
or, as a set, the union of three intervals:
$$(-infty,-5]cup(-3,3)cup[5,+infty). $$
answered 9 hours ago
BernardBernard
132k7 gold badges43 silver badges126 bronze badges
$endgroup$
add a comment |
$begingroup$
The equivalences $$lvert Arvert< BiffbegincasesBge 0\ A> -B\ A< Bendcases\ lvert Arvertge Biff Ble 0lor begincasesB> 0\ Ale -BendcaseslorbegincasesB>0\ Age Bendcases$$
are a good place to start from.
answered 9 hours ago
Gae. S.Gae. S.
2,0648 silver badges17 bronze badges
$endgroup$
$begingroup$
In your first equivalence, the inequalities should be strict.
$endgroup$
– Bernard
8 hours ago
$begingroup$
@Bernard Yes, thank you.
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
Even $B$ should be positive, since $|A|ge 0$.
$endgroup$
– Bernard
8 hours ago
$begingroup$
@Bernard For $B=0$ the other two inequations sort it out (though you may argue that so do they when $B<0$).
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
There remains that $|A|<BRightarrow B>0$, from a strictly logical point of view.
$endgroup$
– Bernard
8 hours ago
|
show 2 more comments
$begingroup$
For $|x|<3$, the LHS is negative and the inequation certainly holds. Now assuming $|x|>3$, we can solve the equation
$$frac1x=frac12,$$ which gives $|x|=5$. The inequation holds for $|x|ge5$. Your resolution was right.
So the solution set is
$$(-3,3)cup(-infty,-5]cup[5,infty).$$
answered 9 hours ago
Yves DaoustYves Daoust
145k10 gold badges89 silver badges248 bronze badges
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Anyway, in your case, $yge 0$, so you should transform all this to
$$|x|<3;textit or ; |x|ge 5iff -3<x<3;textit or ; xge 5;textit or ; xle -5$$
or, as a set, the union of three intervals:
$$(-infty,-5]cup(-3,3)cup[5,+infty). $$
answered 9 hours ago
BernardBernard
132k7 gold badges43 silver badges126 bronze badges
$endgroup$
add a comment |
$begingroup$
Anyway, in your case, $yge 0$, so you should transform all this to
$$|x|<3;textit or ; |x|ge 5iff -3<x<3;textit or ; xge 5;textit or ; xle -5$$
or, as a set, the union of three intervals:
$$(-infty,-5]cup(-3,3)cup[5,+infty). $$
answered 9 hours ago
BernardBernard
132k7 gold badges43 silver badges126 bronze badges
$endgroup$
add a comment |
$begingroup$
Anyway, in your case, $yge 0$, so you should transform all this to
$$|x|<3;textit or ; |x|ge 5iff -3<x<3;textit or ; xge 5;textit or ; xle -5$$
or, as a set, the union of three intervals:
$$(-infty,-5]cup(-3,3)cup[5,+infty). $$
answered 9 hours ago
BernardBernard
132k7 gold badges43 silver badges126 bronze badges
$endgroup$
Anyway, in your case, $yge 0$, so you should transform all this to
$$|x|<3;textit or ; |x|ge 5iff -3<x<3;textit or ; xge 5;textit or ; xle -5$$
or, as a set, the union of three intervals:
$$(-infty,-5]cup(-3,3)cup[5,+infty). $$
answered 9 hours ago
BernardBernard
132k7 gold badges43 silver badges126 bronze badges
answered 9 hours ago
BernardBernard
132k7 gold badges43 silver badges126 bronze badges
answered 9 hours ago
BernardBernard
132k7 gold badges43 silver badges126 bronze badges
answered 9 hours ago
BernardBernard
132k7 gold badges43 silver badges126 bronze badges
132k7 gold badges43 silver badges126 bronze badges
add a comment |
add a comment |
$begingroup$
The equivalences $$lvert Arvert< BiffbegincasesBge 0\ A> -B\ A< Bendcases\ lvert Arvertge Biff Ble 0lor begincasesB> 0\ Ale -BendcaseslorbegincasesB>0\ Age Bendcases$$
are a good place to start from.
answered 9 hours ago
Gae. S.Gae. S.
2,0648 silver badges17 bronze badges
$endgroup$
$begingroup$
In your first equivalence, the inequalities should be strict.
$endgroup$
– Bernard
8 hours ago
$begingroup$
@Bernard Yes, thank you.
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
Even $B$ should be positive, since $|A|ge 0$.
$endgroup$
– Bernard
8 hours ago
$begingroup$
@Bernard For $B=0$ the other two inequations sort it out (though you may argue that so do they when $B<0$).
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
There remains that $|A|<BRightarrow B>0$, from a strictly logical point of view.
$endgroup$
– Bernard
8 hours ago
|
show 2 more comments
$begingroup$
The equivalences $$lvert Arvert< BiffbegincasesBge 0\ A> -B\ A< Bendcases\ lvert Arvertge Biff Ble 0lor begincasesB> 0\ Ale -BendcaseslorbegincasesB>0\ Age Bendcases$$
are a good place to start from.
answered 9 hours ago
Gae. S.Gae. S.
2,0648 silver badges17 bronze badges
$endgroup$
$begingroup$
In your first equivalence, the inequalities should be strict.
$endgroup$
– Bernard
8 hours ago
$begingroup$
@Bernard Yes, thank you.
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
Even $B$ should be positive, since $|A|ge 0$.
$endgroup$
– Bernard
8 hours ago
$begingroup$
@Bernard For $B=0$ the other two inequations sort it out (though you may argue that so do they when $B<0$).
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
There remains that $|A|<BRightarrow B>0$, from a strictly logical point of view.
$endgroup$
– Bernard
8 hours ago
|
show 2 more comments
$begingroup$
The equivalences $$lvert Arvert< BiffbegincasesBge 0\ A> -B\ A< Bendcases\ lvert Arvertge Biff Ble 0lor begincasesB> 0\ Ale -BendcaseslorbegincasesB>0\ Age Bendcases$$
are a good place to start from.
answered 9 hours ago
Gae. S.Gae. S.
2,0648 silver badges17 bronze badges
$endgroup$
The equivalences $$lvert Arvert< BiffbegincasesBge 0\ A> -B\ A< Bendcases\ lvert Arvertge Biff Ble 0lor begincasesB> 0\ Ale -BendcaseslorbegincasesB>0\ Age Bendcases$$
are a good place to start from.
answered 9 hours ago
Gae. S.Gae. S.
2,0648 silver badges17 bronze badges
edited 8 hours ago
answered 9 hours ago
Gae. S.Gae. S.
2,0648 silver badges17 bronze badges
answered 9 hours ago
Gae. S.Gae. S.
2,0648 silver badges17 bronze badges
answered 9 hours ago
Gae. S.Gae. S.
2,0648 silver badges17 bronze badges
2,0648 silver badges17 bronze badges
$begingroup$
In your first equivalence, the inequalities should be strict.
$endgroup$
– Bernard
8 hours ago
$begingroup$
@Bernard Yes, thank you.
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
Even $B$ should be positive, since $|A|ge 0$.
$endgroup$
– Bernard
8 hours ago
$begingroup$
@Bernard For $B=0$ the other two inequations sort it out (though you may argue that so do they when $B<0$).
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
There remains that $|A|<BRightarrow B>0$, from a strictly logical point of view.
$endgroup$
– Bernard
8 hours ago
|
show 2 more comments
$begingroup$
In your first equivalence, the inequalities should be strict.
$endgroup$
– Bernard
8 hours ago
$begingroup$
@Bernard Yes, thank you.
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
Even $B$ should be positive, since $|A|ge 0$.
$endgroup$
– Bernard
8 hours ago
$begingroup$
@Bernard For $B=0$ the other two inequations sort it out (though you may argue that so do they when $B<0$).
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
There remains that $|A|<BRightarrow B>0$, from a strictly logical point of view.
$endgroup$
– Bernard
8 hours ago
$begingroup$
In your first equivalence, the inequalities should be strict.
$endgroup$
– Bernard
8 hours ago
$begingroup$
In your first equivalence, the inequalities should be strict.
$endgroup$
– Bernard
8 hours ago
$begingroup$
@Bernard Yes, thank you.
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
@Bernard Yes, thank you.
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
Even $B$ should be positive, since $|A|ge 0$.
$endgroup$
– Bernard
8 hours ago
$begingroup$
Even $B$ should be positive, since $|A|ge 0$.
$endgroup$
– Bernard
8 hours ago
$begingroup$
@Bernard For $B=0$ the other two inequations sort it out (though you may argue that so do they when $B<0$).
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
@Bernard For $B=0$ the other two inequations sort it out (though you may argue that so do they when $B<0$).
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
There remains that $|A|<BRightarrow B>0$, from a strictly logical point of view.
$endgroup$
– Bernard
8 hours ago
$begingroup$
There remains that $|A|<BRightarrow B>0$, from a strictly logical point of view.
$endgroup$
– Bernard
8 hours ago
|
show 2 more comments
$begingroup$
For $|x|<3$, the LHS is negative and the inequation certainly holds. Now assuming $|x|>3$, we can solve the equation
$$frac1x=frac12,$$ which gives $|x|=5$. The inequation holds for $|x|ge5$. Your resolution was right.
So the solution set is
$$(-3,3)cup(-infty,-5]cup[5,infty).$$
answered 9 hours ago
Yves DaoustYves Daoust
145k10 gold badges89 silver badges248 bronze badges
$endgroup$
add a comment |
$begingroup$
For $|x|<3$, the LHS is negative and the inequation certainly holds. Now assuming $|x|>3$, we can solve the equation
$$frac1x=frac12,$$ which gives $|x|=5$. The inequation holds for $|x|ge5$. Your resolution was right.
So the solution set is
$$(-3,3)cup(-infty,-5]cup[5,infty).$$
answered 9 hours ago
Yves DaoustYves Daoust
145k10 gold badges89 silver badges248 bronze badges
$endgroup$
add a comment |
$begingroup$
For $|x|<3$, the LHS is negative and the inequation certainly holds. Now assuming $|x|>3$, we can solve the equation
$$frac1x=frac12,$$ which gives $|x|=5$. The inequation holds for $|x|ge5$. Your resolution was right.
So the solution set is
$$(-3,3)cup(-infty,-5]cup[5,infty).$$
answered 9 hours ago
Yves DaoustYves Daoust
145k10 gold badges89 silver badges248 bronze badges
$endgroup$
For $|x|<3$, the LHS is negative and the inequation certainly holds. Now assuming $|x|>3$, we can solve the equation
$$frac1x=frac12,$$ which gives $|x|=5$. The inequation holds for $|x|ge5$. Your resolution was right.
So the solution set is
$$(-3,3)cup(-infty,-5]cup[5,infty).$$
answered 9 hours ago
Yves DaoustYves Daoust
145k10 gold badges89 silver badges248 bronze badges
edited 7 hours ago
answered 9 hours ago
Yves DaoustYves Daoust
145k10 gold badges89 silver badges248 bronze badges
answered 9 hours ago
Yves DaoustYves Daoust
145k10 gold badges89 silver badges248 bronze badges
answered 9 hours ago
Yves DaoustYves Daoust
145k10 gold badges89 silver badges248 bronze badges
145k10 gold badges89 silver badges248 bronze badges
add a comment |
add a comment |
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