Order in the real numbers
The real numbers possess a crucial property of order, allowing for the comparison of two numbers to determine which is greater or if they are equal We define a subset of real numbers, denoted as P, representing positive numbers, where we express this relationship symbolically as x > 0 if x belongs to P Additionally, we will consider three essential properties related to this set.
Property 1.1.1 Every real numberxhas one and only one of the following prop- erties:
Property 1.1.2 Ifx,y∈P, thenx+y∈P (in symbolsx >0,y >0⇒x+y >0).
Property 1.1.3 If x,y∈P, thenxy∈P (in symbols x >0,y >0⇒xy >0).
The real line is a geometric representation of real numbers, where the number "0" divides it into two parts: positive numbers on the right, including "1." The number "1" is classified as positive, as it satisfies the property that for any number x, 1 + x = x, which implies that if it were negative, it would lead to contradictions regarding the classification of numbers.
In mathematical terms, we define the relation \( a > b \) if \( a - b \in P \), indicating that \( a \) is greater than \( b \) Conversely, \( a < b \) is defined by \( b - a \in P \), meaning \( a \) is smaller than \( b \) It is important to note that \( a < b \) is equivalent to \( b > a \) Additionally, we can express that \( a \) is smaller than or equal to \( b \) with the notation \( a \leq b \), which holds true if either \( a < b \) or \( a = b \).
We will denote byRthe set of real numbers and byR + the setP of positive real numbers.
Example 1.1.4 (i) If a < bandc is any number, then a+c < b+c.
(ii) If a < bandc >0, thenac < bc.
In fact, to prove (i) we see that a+c < b+c ⇔ (b+c)−(a+c) >0 ⇔ b−a >0⇔a < b To prove (ii), we proceed as follows:a < b⇒b−a >0 and sincec >0, then (b−a)c >0, thereforebc−ac >0 and thenac < bc.
Exercise 1.1 Given two numbersaandb, exactly one of the following assertions is satisfied,a=b,a > bora < b.
Exercise 1.2 Prove the following assertions.
(ii) Ifb >0, we have that a b >1 if and only ifa > b.
The absolute valueof a real numberx, which is denoted by|x|, is defined as
Geometrically,|x|is the distance of the numberx(on the real line) from the origin
0 Also,|a−b| is the distance between the real numbersaandb on the real line.
Exercise 1.4 For any real numbersx, aandb, the following hold.
(i) |x| ≥0, and is equal to zero only whenx= 0.
Proposition 1.1.5 (Triangle inequality) The triangle inequality states that for any pair of real numbersa andb,
|a+b| ≤ |a|+|b|. Moreover, the equality holds if and only if ab≥0.
Proof Both sides of the inequality are positive; then using Exercise 1.3 it is suffi- cient to verify that|a+b| 2 ≤(|a|+|b|) 2 :
In the previous relations we observe only one inequality, which is obvious since ab≤ |ab| Note that, whenab≥0, we can deduce thatab=|ab|=|a| |b|, and then the equality holds
Thegeneral form of the triangle inequality for real numbersx 1,x 2 , , x n , is
The equality is maintained when all \( x_i \) values share the same sign, which can be demonstrated through similar reasoning or by induction A frequently utilized variation of the previous inequality is presented below.
Exercise 1.5 Letx,y,a,b be real numbers, prove that
Exercise 1.6 For real numbersa,b,c, prove that
Exercise 1.7 Leta,b be real numbers such that 0≤a≤b≤1 Prove that
Exercise 1.8 Prove that ifn,mare positive integers, then m n f(a, b, c, d)> f(a, b, d, c).
Exercise 1.13 (IMO, 1960) For which real values of x the following inequality holds:
Exercise 1.14 Prove that for any positive integern, the fractional part of√
Exercise 1.15 (Short list IMO, 1996) Let a, b, c be positive real numbers such thatabc= 1 Prove that ab a 5 +b 5 +ab + bc b 5 +c 5 +bc+ ca c 5 +a 5 +ca ≤1.
The quadratic function ax 2 + 2bx + c
The inequality \( x^2 \geq 0 \) is a fundamental principle for all real numbers, allowing us to derive numerous other inequalities This property is particularly useful in determining the maximum or minimum values of quadratic functions in the form \( ax^2 + 2bx + c \) Quadratic functions frequently arise in optimization problems and inequalities, making this inequality essential for mathematical analysis and problem-solving.
One common example consists in proving that ifa >0, the quadratic function ax 2 + 2bx+c will have its minimum atx=− a b and the minimum value isc− b a 2
In fact, ax 2 + 2bx+c=a x 2 + 2b ax+b 2 a 2
≥ 0 and the minimum value of this expression, zero, is attained when x=− a b , we conclude that the minimum value of the quadratic function is c− b a 2
Ifa a l , then let us write a k =ρ+τ, a l =ρ−τ (0< τ ≤ρ).
If now 0≤σ < τ≤ρ, then aT-transformation is defined by
In the context of aT-transformation, we denote the relationship where b arises from a as b = T a This definition does not require that either a or b be arranged in a decreasing order To demonstrate the sufficiency of our comparability condition, we need to establish two key lemmas.
Lemma 1.8.3 Ifb=T a, then[b]≤[a]with equality taking place only when all the x i ’s are equal.
Proof We may rearrange (a) and (b) so thatk= 1,l= 2 Thus
!(x 1 x 2) ρ−τ x a 3 3 ã ã ãx a n n (x τ+σ 1 −x τ+σ 2 )(x τ−σ 1 −x τ−σ 2 )≥0 with equality being the case only when all thex i ’s are equal
Lemma 1.8.4 If (b)≺(a), but(b)is not identical to(a), then (b)can be derived from(a)using the successive application of a finite number ofT-transformations.
The discrepancy between two sets, denoted as ν - bν, represents the count of non-zero differences If this discrepancy equals zero, it indicates that the sets are identical We will demonstrate the lemma through induction, starting with the assumption that it holds true for discrepancies less than r, and then proving its validity for discrepancies equal to r.
Suppose then that (b) ≺ (a) and that the discrepancy is r > 0 Since n i=1 a i = n i=1 b i , and
In the equation (a ν − b ν ) = 0, not all differences can be zero, indicating the presence of both positive and negative differences The first non-zero difference must be positive due to the condition (b)≺(a) Consequently, we can identify indices k and l such that b k < a k, b k +1 = a k +1, , b l−1 = a l−1, and b l > a l This means that a l − b l represents the first negative difference, while a k − b k is the last positive difference that occurs before it.
We takea k =ρ+τ,a l =ρ−τ, and defineσby σ= max(|b k −ρ|,|b l −ρ|).
Then 0 < τ ≤ ρ, since a k > a l Also, one (possible both) of b l −ρ = −σ or b k −ρ=σ is true, since b k ≥ b l , and σ < τ, since b k < a k and b l > a l Hence
We define the variables as follows: \( a_k = \rho + \sigma \), \( a_l = \rho - \sigma \), and \( a_\nu = a_\nu \) (where \( \nu \) can be either \( k \) or \( l \)) If \( b_k - \rho = \sigma \), then \( a_k = b_k \); conversely, if \( b_l - \rho = -\sigma \), then \( a_l = b_l \) Each pair \( (a_k, b_k) \) and \( (a_l, b_l) \) reduces the discrepancy \( r \) between \( (b) \) and \( (a) \) by one unit, resulting in a smaller discrepancy of \( r - 1 \) or \( r - 2 \).
Next, comparing the definition of (a ) with the definition of theT-transfor- mation, and observing that 0≤σ < τ ≤ρ, we can infer that (a ) arises from (a) by aT-transformation.
To establish that (b)≺(a), we need to confirm that both conditions of the relation ≺ are met, while also ensuring that the order of (a) is non-increasing Specifically, we find that the expression k + a l equals 2ρ, which is equivalent to a k + a l, alongside the summations n i=1 b i, n i=1 a i, and n i=1 a i.
For the second one, we must prove that b 1+b 2+ã ã ã+b ν ≤α 1 +α 2 +ã ã ã+α ν
Now, this is true ifν < korν ≥l, as can be established by using the definition of
The second condition of (b)≺(a) holds true for ν = k, as it is valid for ν = k−1 with the condition b k ≤ a k Additionally, for values where k < ν < l, the condition remains valid since ν = k and the intervening bands a are identical.
Finally, we observe that b k ≤ρ+|b k −ρ| ≤ρ+σ=a k , b l ≥ρ− |b l −ρ| ≥ρ−σ=a l , and then, using (1.12), a k− 1 =a k− 1 ≥a k =ρ+τ > ρ+σ=a k ≥b k ≥b k+1 =a k+1=a k+1 , a l− 1 =a l− 1=b l− 1 ≥b l ≥a l =ρ−σ > ρ−τ =a l ≥a l+1=a l +1
The inequalities involving a are as required.
We have thus proved that (b)≺(a ), a set arising from (a) using a transfor- mationT and having a discrepancy from (b) of less thanr This proves the lemma and completes the proof of Muirhead’s theorem
Muirhead's theorem illustrates that the difference between two comparable means can be expressed as a sum of positive terms through the repeated application of the T-transformation This result provides a new proof for the Arithmetic Mean-Geometric Mean (AM-GM) inequality.
Example 1.8.5 (The AM-GM inequality) For real positive numbers y 1 , y 2 , , y n , y 1+y 2+ã ã ã+y n n ≥ √ n y 1 y 2 ã ã ãy n
Note that the AM-GM inequality is equivalent to
By Muirhead’s theorem we can show that
We present an alternative proof of the AM-GM inequality, drawing on concepts from the proof of Muirhead’s theorem to demonstrate its application effectively.
Since (x ν r −x ν s )(x r −x s )>0, unlessx r =x s , the inequality follows.
Example 1.8.6 If a,b are positive real numbers, then a 2 b + b 2 a ≥√ a+√ b.
Setting x=√ a,y=√ band simplifying, we have to prove x 3 +y 3 ≥xy(x+y).
Using Muirhead’s theorem, we get
2xy(x+y) = [2,1], and thus the result follows.
Example 1.8.7 If a,b,c are non-negative real numbers, prove that a 3 +b 3 +c 3 +abc≥ 1
It is not difficult to see that
Then we need to prove that
This follows using the inequalities [3,0,0]≥[2,1,0] and [1,1,1]≥0.
Example 1.8.8 If a,b,c are non-negative real numbers, prove that a+b+c≤ a 2 +b 2
The inequalities are equivalent to the following:
2(a 2 bc+ab 2 c+abc 2 )≤ab(a 2 +b 2 ) +bc(b 2 +c 2 ) +ca(c 2 +a 2 )≤2(a 4 +b 4 +c 4 ), which is in turn equivalent to [2,1,1]≤[3,1,0]≤[4,0,0] Using Muirhead’s theo- rem we arrive at the result.
Exercise 1.115 Prove that any three positive real numbersa,b andcsatisfy a 5 +b 5 +c 5 ≥a 3 bc+b 3 ca+c 3 ab.
Exercise 1.116 (IMO, 1961) Let a, b, c be the lengths of the sides of a triangle, and let (ABC) denote its area Prove that
Exercise 1.117 Leta,b,cbe positive real numbers Prove that a
Exercise 1.118 (IMO, 1964) Leta, b,c be positive real numbers Prove that a 3 +b 3 +c 3 + 3abc≥ab(a+b) +bc(b+c) +ca(c+a).
Exercise 1.119 (Short list Iberoamerican, 2003) Leta, b, cbe positive real num- bers Prove that a 3 b 2 −bc+c 2 + b 3 c 2 −ca+a 2 + c 3 a 2 −ab+b 2 ≥a+b+c.
Exercise 1.120 (Short list IMO, 1998) Let a, b,c be positive real numbers such thatabc= 1 Prove that a 3
Two basic inequalities
This section discusses two fundamental geometric principles related to triangles The first, known as the triangle inequality (D1), establishes essential relationships between the lengths of a triangle's sides The second principle, referred to as D2, highlights a crucial observation: knowing the largest angle in a triangle allows us to identify its longest side.
D1 If A,B andC are points on the plane, then
Moreover, the equality holds if and only ifB lies on the line segmentAC.
D2 In a triangle, the longest side is opposite to the greatest angle and vice versa.
Hence, if in the triangleABC we have∠A >∠B, thenBC > CA.
Exercise 2.1 (i) If a, b, c are positive numbers with a < b+c, b < c+aand c < a+b, then a triangle exists with side lengthsa,b andc.
(ii) To be able to construct a triangle with side lengthsa≤b≤c, it is sufficient thatc < a+b.
(iii) It is possible to construct a triangle with sides of length a, b and c if and only if there are positive numbersx,y,zsuch thata=x+y,b=y+zand c=z+x.
Exercise 2.2 (i) If it is possible to construct a triangle with side-lengthsa < b < c, then it is possible to construct a triangle with side-lengths√ a 1 2 AB.
Exercise 2.5 LetABCD be a convex quadrilateral, prove that
(i) ifAB+BD < AC+CD, thenAB < AC,
(ii) if∠A >∠C and∠D >∠B, thenBC > 1 2 AD.
Exercise 2.6 If a 1, a 2, a 3, a 4 and a 5 are the lengths of the sides of a convex pentagon and ifd 1,d 2,d 3,d 4 andd 5are the lengths of its diagonals, prove that
Exercise 2.7 The lengthm a of the medianAA of a triangleABC satisfiesm a > b + c−a
Exercise 2.8 If the length m a of the median AA of a triangle ABC satisfies m a > 1 2 a, prove that∠BAC