Lesson Proving inequalities

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Proving inequalities


Problem 1

Show that in general,  if  "a   and  "b"  are positive and   a < b,  then   a%2F%28ax+%2B+y%29 < b%2F%28bx+%2B+y%29,
where  x  and  y  can be any positive real numbers.

Solution

The standard way to prove such statements is to start from the given inequality and to transform it to the simplest form.

using EQUIVALENT transformations.


If the final simplified inequality is TRUE, then the starting inequality is TRUE, too, since you can reverse the chain 
of equivalent transformations.


So, we start from  

    a%2F%28ax+%2B+y%29 < b%2F%28bx+%2B+y%29.                   (1)


Multiply both sides by the positive number  (ax+y)*(bx+y).  Then inequality (1) is EQUIVALENT to

    a*(bx+y) < b*(ax+y).               (2)


Inequality (2) is EQUIVALENT to

    abx + ay < abx + by                (3)


which is equivalent to

    ay < by                            (4)


which with positive "y" is equivalent to

    a < b,                             (5)


which is given.


So, the reversed chain of VALID inequalities   (5) ---> (4) ---> (3) ---> (2) ---> (1)  proves the statement.

Problem 2

Let  "a"  and  "b"  be positive real numbers   a >= 0,   b >= 0.   Prove that   a^4 + b^4 >= a^3*b + a*b^3.

Solution

Let's consider this expression

    a^4 - a^3b + b^4 - ab^3.



Transform it this way  

    a^4 - a^3b + b^4 - ab^3 = a%5E3%2A%28a-b%29 + b%5E3%2A%28b-a%29 = %28a-b%29%2A%28a%5E3-b%5E3%29 = %28a-b%29%2A%28a-b%29%2A%28a%5E2+%2B+ab+%2B+b%5E2%29 = %28a-b%29%5E2%2A%28a%5E2%2Bab%2Bb%5E2%29.



So, our starting expression is the product of two quadratic polynomials

    %28a-b%29%5E2  and  a%5E2+%2B+ab+%2B+b%5E2.



They both are positively defined; in other words, they never take negative values.


Therefore,  %28a-b%29%5E2%2A%28a%5E2%2Bab%2Bb%5E2%29 >= 0  for all values of "a" and "b".



It implies that the original expression is never negative 

    a^4 - a^3b + b^4 - ab^3 >= 0.



It means that

    a^4 + b^4 >= a^3b + ab^3,


which is what has to be proved.


At this point, the proof is completed.

Problem 3

Prove that if  |x+3| < 1/2 ,   then  |4x+13| < 3.

Solution

Your starting inequality is

    |x+3| < 1/2.     (1)


Taking off the absolute value symbol, it means that

    -1/2 < x+3 < 1/2.      (the compound inequality)


Multiply the last compound inequality by 4  (multiply all three its terms).
You will get an equivalent inequality

    -2 < 4x + 12 < 2.


Add 1 (one) to the last compound inequality (to all its three terms).
You will get an equivalent inequality

    -1 < 4x + 13 < 3.    (2)


But if (2) is valid, then also

    -3 < 4x + 13 < 3     (3)

is valid, too.  


    +--------------------------------------------------------------------------+
    |   In Math, they say "if (2) is valid, then (3) is valid even more so".   |
    +--------------------------------------------------------------------------+


The last inequality (3)  is the same as 

    |4x+13| < 3,

which we are requested to prove.


At this point, the proof is complete.


My other lessons on solving inequalities are
    - Solving simple and simplest linear inequalities
    - Solving absolute value inequalities
    - Advanced problems on solving absolute value inequalities
    - Solving systems of linear inequalities in one unknown
    - Solving compound inequalities

    - What number is greater? Comparing magnitude of irrational numbers
    - Arithmetic mean and geometric mean inequality
    - Arithmetic mean and geometric mean inequality - Geometric interpretations
    - Harmonic mean
    - Prove that if a, b, and c are the sides of a triangle, then so are sqrt(a), sqrt(b), and sqrt(c)

    - Solving problems on quadratic inequalities
    - Solving inequalities for high degree polynomials factored into a product of linear binomials
    - Solving inequalities for rational functions with numerator and denominator factored into a product of linear binomials
    - Solving inequalities for rational functions with non-zero right side
    - Another way solving inequalities for rational functions with non-zero right side

    - Advanced problems on inequalities
    - Challenging problems on inequalities
    - Solving systems of inequalities in two unknowns graphically in a coordinate plane
    - Solving word problems on inequalities
    - Math circle level problem on inequalities
    - Math Olympiad level problems on inequalities
    - Entertainment problems on inequalities
under the topic  Inequalities, trichotomy of the section  Algebra-I.

My lessons on domains of functions are
    - Domain of a function which is a quadratic polynomial under the square root operator
    - Domain of a function which is a high degree polynomial under the square root operator
    - Domain of a function which is the square root of a rational function.
under the topic  Functions, Domain  of the section  Algebra-I.

See also  OVERVIEW of lessons on inequalities and domains of functions.

Use this file/link  ALGEBRA-I - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-I.



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