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My Notes on Inequalities
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Updated on: 01 Dec 2009, 05:55
Please find my notes on Inequalities. Hope it would be a help to you. Please add if you have anything more
Linear Inequalities An inequality looks like an equation except that the equality symbol = is replaced by one of the inequality symbols <, >, ≤ , or ≥. The solutions of inequalities are, in general, intervals of numbers on the real line instead of just a single number or a small set of numbers.
To solve an inequality that contains a variable means to find all values of the variable for which the inequality is true. Unlike an equation, an inequality usually has infinitely many solutions, which form an interval or a union of intervals on the real line.
Rules of Inequalities
Adding the same quantity to each side of an inequality gives an equivalent inequality
A≤B=>A+C≤B+C
Subtracting the same quantity from each side of an inequality gives an equivalent inequality
A≤B=>AC≤BC
Multiplying each side of an inequality by the same positive quantity gives an equivalent inequality
If C>0, then A≤B=>CA≤CB
Multiplying each side of an inequality by the same negative quantity reverses the direction of the inequality
If C<0, then A≤B=>CA≥CB
Taking reciprocals of each side of an inequality involving positive quantities reverses the direction of the inequality.
0<A≤B=>(1/A)≥(1/B)>0
Inequalities can be added
If A≤B and C≤D, then A+C≤B+D
Solving Inequalities
Linear inequalities involving variables may be solved by using the Rules of Inequalities to reduce them to a form like x≤a, , x≥a, x<a, or x>a. where x is our variable and a is a real number
Ex 1: Solve the inequality 3x1≤5 Adding 1 to each side results in an equivalent inequality 3x≤6 Then multiply each side by (1/3) to get another equivalent inequality (1/3)(3x)≤(1/3)(6)
Since this is equivalent to the original inequality, you can say that any number x satisfies the original inequality if and only if it satisfies x≤2.
Ex2: Solve the inequality 3x<9x+4 First subtract 3x from each side to get an equivalent inequality 0<6x+4 Then subtract 4 from each side to get another equivalent inequality 4<6x Now multiply each side by (1/6) to get a third equivalent inequality (2/3)<x
Since this is equivalent to the original inequality, you can say that any number x satisfies the original inequality if and only if it satisfies x>(2/3)
Now let us move to solving Solving a Pair of Simultaneous Inequalities
Ex3: Solve the inequalities 4≤3x2<13 ( Note that this is the same as saying 4≤3x2 and 3x2<13 )
First, add 2 to each side of each inequality to get 4≤3x and 3x<15 Then multiply each side of each inequality by (1/3) to get the solution (4/3)≤x and x<5
Ex4: Solve the inequalities 2x+1≤4x3≤x+7 Subtract 2x and add 3 to each side of the first inequality to get 4≤4x => 1≤x (multiplying each side by (1/4))
Subtract x and add 3 to each side of the second inequality to get 3x≤10 => x≤((10)/3) (multiplying each side by (1/3))
Thus, the original pair of inequalities is equivalent to the pair 1≤x and x≤((10)/3)
Nonlinear Inequalities
Inequalities involving squares and higher powers of the variable are nonlinear inequalities
Solving Nonlinear Inequalities
a) If necessary, rewrite the inequality so that all nonzero terms appear on one side of the inequality sign b) If the nonzero side of the inequality involves quotients, bring them to a common denominator. c) Factor the nonzero side of the inequality. d) List the intervals determined by factorization e) Make a table or diagram of the signs of each factor on each interval. In the last row of the table, determine the sign of the product (or quotient) of these factors. f) Determine the solution set from the last row of the table. Be sure to check whether the inequality is satisfied by some or all of the endpoints of the intervals (this may happen if the inequality involves ≤ or ≥).
Ex5: Solve the inequality x²5x+6≤0 All the nonzero terms are already on one side of the inequality sign, and there are no quotients, so the first two steps are taken care of. Now factor (x3)(x2)≤0
According to your knowledge of the sign of a product or quotient, you can solve this if you know exactly when each of the factors is positive or negative. Since x3 is negative for x<3 and positive for x>3, and x2 is negative for x<2 and positive for x>2, the problem is reduced to examining what happens in the four cases x<3, x>3, x<2 and x>2, and checking the values x=3 and x=2 as well.
So the intervals you need to examine are 1) x<2 2) 2<x<3 3) x>3 you can easily see that (2,3) is in the solution set and that x<2 and x>3 are not
So the final solution is 2≤x≤3
Absolute Value
a=a if a≥0 and –a if a<0
More generally, the distance from x to a on the number line is given by xa.
Properties of Absolute Value
Let a, b, and c be real numbers, with c>0, and let n be an integer
ab=ab (a/b)=((a)/(b)) if b≠0 a^n=a^n x=c is equivalent to x=±c x<c is equivalent to c<x<c x>c is equivalent to (x<c or x>c)
The Triangle Inequality
If a and b are any real numbers, then a+b ≤ a+b
Solving Equations and Inequalities Involving Absolute Value
Ex6: Solve the equation 2x5=3
This is equivalent to 2x5=3 or 2x5=3 ( By property x=c is equivalent to x=±c ) 2x=8 or 2x=2 x=4 or x=1
Ex7: Solve the inequality (3/(3x+7))≥1 Since the absolute value of any number is nonnegative, 3x+7≥0; as long as 3x+7 not equal to 0, you can multiply each side of the inequality by 3x+7 without affecting the inequality
3≥3x+7
3≤3x+7 and 3x+7≤3 (By property x<c is equivalent to c<x<c) 10≤3x and 3x≤4 ((10)/3)≤x and x≤(4/3) ((10)/3)≤x≤(4/3)
Remember, though, that you had to insist on 3x+7 is not equal to 0. so the value x=(7/3) must be excluded The solution is thus ((10)/3)≤x≤(4/3), x not equal to (7/3)
Some more important results on Inequalities
AMGM inequality: For all positive real numbers a1,a2,a3,a4…. The following equality holds good
(a1+a2+a3+….an)/n ≥ (a1*a2*a3…an)^1/n ; equality occurs when a1=a2=a3=….=an
Weighted AMGM inequality: For all positive real numbers a1,a2,a3,a4…. an. And for all non negative real numbers x1,x2….xn and have sum equal to1.
(a1x1+a2x2+a3x3+….anxn) ≥ (a1^x1*a2^x2*a3^x3…an^xn)
Nesbitt inequality: For all non negative real numbers a, b, c
a/(b+c)+b/(a+c)+c/(a+b) ≥ 3/2
Note: For this section, I will be including some example in latter time
Originally posted by jade3 on 28 Nov 2009, 03:33.
Last edited by jade3 on 01 Dec 2009, 05:55, edited 13 times in total.



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Re: My Notes on Inequalities
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28 Nov 2009, 08:03



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Re: My Notes on Inequalities
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29 Nov 2009, 01:49
Thanks for putting this together. I think an example after each minisection would be really helpful!
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Re: My Notes on Inequalities
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29 Nov 2009, 03:24
bb wrote: Thanks for putting this together. I think an example after each minisection would be really helpful! Thanks bb. That is a good suggestion



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Re: My Notes on Inequalities
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01 Dec 2009, 05:54
Inequalities in geometry
Inequality 1: The measure of an exterior of a triangle is greater than the measure of either remote interior angle.
Inequality 2: If one side of a triangle is longer than a second side, then the angle opposite the first side is larger than the angle opposite the second side.
In triangle ABC: If AC > AB, then Angle[ABC] > Angle[ACB]
Inequality 3: If one angle of a triangle is larger than a second angle, then the side opposite the first angle is longer than the side opposite the second angle.
In triangle ABC: If Angle[ABC] > Angle[ACB], then AB > AC
Inequality 4: The sum of the lengths of any two sides of a triangle is greater than the length of the third side
In a Triangle ABC: AB + BC > AC; AB + AC > BC; AC + BC > AB
Inequality 5: If two sides of one triangle are congruent to two sides of another triangle, but the included angle of the first triangle is larger than the included angle of the second then the third side of the first triangle is longer than the third side of the second triangle
In Triangle ABC and Triangle DEF; If AB =DE; BC = EF ;and Angle[ABC]>Angle[DEF], then AC > DF
Inequality 6: If two sides of one triangle are congruent to two sides of another triangle, but the third side of the first triangle is longer than the thirds side of the second, then the included angle of the first triangle is larger than the included angle of the second
In Triangle ABC and Triangle DEF; If BA = ED; BC = EF; AC > DF,then Angle[ABC]> Angle[DEF]
Inequality 7: Of all the plane figures with a given perimeter, the circle has the greatest area.
Inequality 8: Of all the plane figures with a given area, the circle has the least perimeter
Inequality 9: Of all the solids with a given surface area, the sphere has the greatest volume.
Inequality 10: Of all the solids with a given volume, the sphere has the least surface area.
Inequality 11: Of all the triangles with a common base and perimeter, the isosceles triangle has the greatest area.
Inequality 12: Of all the triangles with a common base and area, the isosceles triangle has the smallest perimeter
Inequality 13: Of all the triangles with a given area, the equilateral triangle has the least perimeter
Inequality 14: Of all the triangles with a given perimeter, the equilateral triangle has the greatest area.
Inequality 15: Of all the ngons inscribed in a given circle, the regular ngon has the greatest area.
Inequality 16: Of all the quadrilaterals with a given area, the square has the least perimeter.
Inequality 17: A quadrilateral with given sides has the greatest area when it can be inscribed in a circle.
Note: For this section, I will be including some example in latter time



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Re: My Notes on Inequalities
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10 Jan 2010, 23:11
jade3 wrote: Solving Inequalities
Ex2: Solve the inequality 3x<9x+4 First subtract 3x from each side to get an equivalent inequality 0<6x+4 Then subtract 4 from each side to get another equivalent inequality 4<6x Now multiply each side by (1/6) to get a third equivalent inequality (2/3)<x
Since this is equivalent to the original inequality, you can say that any number x satisfies the original inequality if and only if it satisfies x>(2/3)
I have this odd feeling that I shouldn't subtract (3x) from both sides because I don't know whether (x) is positive or negative. This is because the direction of the signs change when multiplying or dividing by a negative. So I'd leave the equation in its original form for a DS question. Any thoughts?
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Re: My Notes on Inequalities
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25 Aug 2014, 19:04
ael719 wrote: jade3 wrote: Solving Inequalities
Ex2: Solve the inequality 3x<9x+4 First subtract 3x from each side to get an equivalent inequality 0<6x+4 Then subtract 4 from each side to get another equivalent inequality 4<6x Now multiply each side by (1/6) to get a third equivalent inequality (2/3)<x
Since this is equivalent to the original inequality, you can say that any number x satisfies the original inequality if and only if it satisfies x>(2/3)
I have this odd feeling that I shouldn't subtract (3x) from both sides because I don't know whether (x) is positive or negative. This is because the direction of the signs change when multiplying or dividing by a negative. So I'd leave the equation in its original form for a DS question. Any thoughts? Adding or subtracting a variable on both sides of inequalities is always fine as there will not be a change in the sign of inequality. However, we cannot multiply or divide a variable on both sides as there is a possibility of change in the inequality.



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My Notes on Inequalities
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25 Jan 2016, 04:43
jade3 wrote: Ex3: Solve the inequalities 4≤3x2<13 ( Note that this is the same as saying 4≤3x2 and 3x2<13 )
First, add 2 to each side of each inequality to get 4≤3x and 3x<15 Then multiply each side of each inequality by (1/3) to get the solution (4/3)≤x and x<5
4≤3x2 gives 6≤3x No ? jade3 wrote: Ex4: Solve the inequalities 2x+1≤4x3≤x+7 Subtract 2x and add 3 to each side of the first inequality to get 4≤4x => 1≤x (multiplying each side by (1/4)) 4≤2x instead ?



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Re: My Notes on Inequalities
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26 Jan 2016, 11:55
you are right! I also identified the same typo error! boo2gom wrote: jade3 wrote: Ex3: Solve the inequalities 4≤3x2<13 ( Note that this is the same as saying 4≤3x2 and 3x2<13 )
First, add 2 to each side of each inequality to get 4≤3x and 3x<15 Then multiply each side of each inequality by (1/3) to get the solution (4/3)≤x and x<5
4≤3x2 gives 6≤3x No ? jade3 wrote: Ex4: Solve the inequalities 2x+1≤4x3≤x+7 Subtract 2x and add 3 to each side of the first inequality to get 4≤4x => 1≤x (multiplying each side by (1/4)) 4≤2x instead ?



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Re: My Notes on Inequalities
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23 Sep 2016, 18:14
bb wrote: Thanks for putting this together. I think an example after each minisection would be really helpful! Hi was just wondering ..does the GMAT club math book have an inequalities section coz I couldn't find it in the PDF. Thanks Sent from my ONE A2003 using GMAT Club Forum mobile app



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Re: My Notes on Inequalities
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24 Sep 2016, 01:31



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Re: My Notes on Inequalities
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25 Sep 2016, 22:47
Bunuel wrote: Thankyou so much will check them out now! Sent from my ONE A2003 using GMAT Club Forum mobile app



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