My Notes on Inequalities
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Updated on: 01 Dec 2009, 04:55
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=>A-C≤B-C
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 3x-1≤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≤3x-2<13 ( Note that this is the same as saying 4≤3x-2 and 3x-2<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≤4x-3≤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 x²-5x+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 (x-3)(x-2)≤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 x-3 is negative for x<3 and positive for x>3, and x-2 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 |x-a|.
Properties of Absolute Value
Let a, b, and c be real numbers, with c>0, and let n be an integer
|ab|=|a||b|
|(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 |2x-5|=3
This is equivalent to 2x-5=3 or 2x-5=-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
AM-GM 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 AM-GM 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, 02:33.
Last edited by
jade3 on 01 Dec 2009, 04:55, edited 13 times in total.