[3] Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, Inequalities with Beautiful Solutions , GIL Publishing house, 2009. [4] Pham Kim Hung, Secrets in Inequality , GIL Publishing house, 2[r]
(1)USING SIMPLE INEQUALITIES TO SOLVE SOME BEAUTIFUL PROBLEMS
Nguyen Dang Khoa - Hung Vuong High school for the gifted - Phu Tho March 2019
1 Introduction
In the inequality world, there are many simple but effective inequalities, which are very useful for some cases Today, I bring the readers an inequality, and I think it is helpful for you in the future Although I wrote this article very carefully, limited knowledge and language description of the author can lead to some unavoidable mistakes Remember your feedbacks and recommendations are always welcomed, so don’t be hesitant!
Now, let’s start to find incredible things! First, we have a basic inequality
• For any non-negative real numbers a, b, c, we obtain
a2
a2+ 2bc +
b2
b2+ 2ac +
c2
c2+ 2ab ≥1 (∗)
It is simple inequality and we know solution by applying Cauchy - Schwarz inequality
a2 a2+ 2bc +
b2 b2+ 2ac +
c2 c2+ 2ab ≥
(a+b+c)2
a2+ 2bc+b2+ 2ac+c2+ 2ab =
(a+b+c)2 (a+b+c)2 =
We can rewrite that inequality as the following form
bc a2+ 2bc +
ca b2+ 2ac+
ab
c2+ 2ab ≤1
(2)Now, We replace (a, b, c)−→
1
a,
1
b,
1
c
and we have the inequality (∗) becomes
• For any non-negative real numbers a, b, c, we have the inequality
bc
2a2+bc +
ca
2b2+ca +
ab
2c2+ab ≥1 (∗∗)
or
a2
2a2+bc +
b2
2b2+ca+
c2
2c2+ab ≤1
The inequality (∗∗) is also proved byCauchy - Schwarz inequality, but the thing which author want to show you is relation between two inequalities (∗) and (∗∗)
From those inequalities, we will discover, extend and some beautiful problems
2 Examples and problems
2.1 Examples
First, we can change the inequality (∗) is equivalent to
a2
a2+ 2bc +
b2
b2+ 2ac +
c2
c2+ 2ab ≥1 or
a3
a3+ 2abc +
b3
b3+ 2abc +
c3
c3+ 2abc ≥1
We see denominators of those fractions have 2abc, if we let abc= then we have inequality is equivalent to
a3
a3+ 2 +
b3
b3 + 2 +
c3
c3+ 2 ≥1
or
1
a3+ 2 +
1
b3 + 2 +
1
c3+ 2 ≤1
So we have example
Example Given positive real numbers a, b, csuch that abc= Prove that
1
a3+ 2 +
1
b3 + 2 +
1
c3+ 2 ≤1
Since abc= then we also have inequality
1
a+ +
b+ +
(3)Proof Sine abc= then there exists x, y, z are positive such that a= x
2
yz, b= y2 zx, c=
z2 xy
So the inequality becomes
yz x2+ 2yz +
zx y2+ 2zx+
xy
z2+ 2xy ≤1
It is inequality (∗), so we are done
Example Given positive real numbers a, b, csatisfying abc= Show that
1 2a+ +
1 2b+ +
1
2c+ ≥1
Proof We will solve this example like example Let a= x
2
yz, b = y2 zx,c=
z2
xy (x, y, z >0)
The inequality becomes
yz
2x2+yz +
zx
2y2+zx+
xy
2z2 +xy ≥1
This is inequality (∗∗), we are done
Now, combining example and example 2, we have a new example
Example Let a, b, care positive real numbers satisfying abc= Show that
2a+ + 2b+ +
1 2c+ ≥
1
a+ +
b+ +
c+ and
a a+ +
b b+ +
c c+ ≥
a
2a+ +
b
2b+ +
c
2c+
Example (Nguyen Dang Khoa)Given positive real numbersa,b,csuch thata2+b2+c2 = 1.
Prove the inequality
a a2+ 2bc
2
+
b b2+ 2ca
2
+
c c2+ 2ab
2
≥1
We can see this inequality has square of fraction so we always think use inequality 3(a2+b2+c2)≥(a+b+c)2 But if we use it, we have
a a2+ 2bc
2
+
b b2+ 2ca
2
+
c c2 + 2ab
2
≥
a a2+ 2bc +
b b2+ 2ca +
c c2+ 2ab
2
(4)So it is enough to show that a
a2+ 2bc+
b b2+ 2ca+
c c2+ 2ab ≥
√
3
But it is hard inequality, we can use S.O.S method to prove it Is that the author mean? Now, let consider a beautiful solution
Proof Applying Cauchy-Schwarz inequality and inequality (∗) we have
"
a a2+ 2bc
2
+
b b2+ 2ca
2
+
c c2+ 2ab
2#
(a2+b2+c2)
≥
a2 a2+ 2bc +
b2 b2+ 2ac +
c2 c2+ 2ab
2
≥1 Combining assumption a2+b2+c2 = 1, we get
a a2+ 2bc
2
+
b b2+ 2ca
2
+
c c2+ 2ab
2
≥1 With the same idea, we have another inequality
a2
(a2+ 2bc)3 +
b2
(b2+ 2ac)3 +
c2
(c2+ 2ab)3 ≥1
We can use Holder inequality to prove it We have
a2
(a2+ 2bc)3 +
b2
(b2+ 2ac)3 +
c2
(c2 + 2ab)3
(a2+b2+c2)(a2+b2 +c2)
≥
a2
a2+ 2bc +
b2
b2+ 2ac +
c2
c2+ 2ab
3
≥1
It implies a
2
(a2+ 2bc)3 +
b2
(b2+ 2ac)3 +
c2
(c2+ 2ab)3 ≥1
From that examples, we have the general problem
General problem Given positive real numbers a, b, c such that a2 +b2 +c2 = and k is positive integer Prove that
a2
(a2+ 2bc)k +
b2
(b2+ 2ca)k +
c2
(c2+ 2ab)k ≥1
Proof Note thata2+ 2bc≤a2+b2+c2 = 1, it implies (a2+ 2bc)k≤a2 + 2bc.
So we obtain
a2
(a2+ 2bc)k +
b2
(b2+ 2ca)k +
c2
(c2+ 2ab)k ≥
a2
a2+ 2bc +
b2
b2+ 2ca +
c2
(5)This solution is very simple, it shows that for some cases general problem is not harder than original problem
Example (Nguyen Dang Khoa) Given the positive real numbers a, b, chave the sum is Show that
a
3bc+a + b
3ca+b + c
3ab+c ≥
3
First, we have assumption is a+b+c= so we have idea is multiple a+b+cor use directly
Cauchy - Schwarz inequality But this method is not really useful So we need more skillful in some steps
Proof We rewrite the inequality in the form
bc
3bc+a + ca
3ca+b + ab
3ab+c ≤
1
Here, we can use a+b+c= and applyCauchy - Schwarz inequality, we have
bc
3bc+a =
bc
3bc+a(a+b+c) =
bc
(a2+ 2bc) + (ab+bc+ca) ≤
1
bc a2+ 2bc +
bc ab+bc+ca
Simillarly, we constitute the different inequalities and we get
bc
3bc+a + ca
3ca+b + ab
3ab+c ≤
1
X ab
2ab+c2 +
X ab
ab+bc+ca
≤
2 So we are done The equality occurs if a=b=c=
3
Example (Nguyen Dang Khoa)Prove this following inequality witha,b,care non-negative real numbers
bc
2a2 + (b+c)2 +
ca
2b2+ (c+a)2 +
ab
2c2 + (a+b)2 ≤
1
Proof Applying Cauchy - Schwarz inequality, we obtain
bc
2a2+ (b+c)2 =
bc
2a2+b2+ 2bc+c2 =
bc
a2+ 2bc+ (a2+b2+c2) ≤
1
bc a2+ 2bc +
bc a2+b2+c2
We constitute other inequalities, we have
X bc
2a2 + (b+c)2 ≤
1
bc a2+ 2bc +
ca b2+ 2ca+
ab c2+ 2ab+
ab+bc+ca a2+b2+c2
It is easy to see that ab+bc+ca
(6)bc
2a2 + (b+c)2 +
ca
2b2+ (c+a)2 +
ab
2c2 + (a+b)2 ≤
1
The equality holds if and only if a=b=cor a= 0, b=c orb = 0, a=cor c= 0, a=b
Now, we change that inequality in the form
2a2+b2+c2
2a2+ (b+c)2 +
2b2+a2+c2
2b2+ (b+c)2 +
2c2+a2+b2
2c2+ (a+b)2 ≥2
Let a2+b2+c2 = then we have a new inequality
Example 6.1 Given the positive real numbera ,b, csastisfying a2+b2+c2 = Prove that
a2+ 1
a2 + 2bc+ 1 +
b2+ 1
b2+ 2ca+ 1 +
c2+ 1
c2+ 2ab+ 1 ≥2
Example (Nguyen Dang Khoa) Prove the following inequality witha,b,care positive real numbers
a2 +bc
a2+ 2bc+
b2+ac
b2+ 2ac+
c2+ab
c2+ 2ab ≥2
Proof The original inequality is eqivalent to 2(a2+bc)
a2+ 2bc +
2(b2+ac)
b2+ 2ac +
2(c2+ab)
c2+ 2ab ≥4
or
2(a2+bc)
a2+ 2bc −1 +
2(b2+ac)
b2+ 2ac −1 +
2(c2+ab)
c2+ 2ab −1≥1
or
a2
a2+ 2bc+
b2
b2+ 2ca+
c2
c2+ 2ab ≥1
This is inequality (∗), so that inequality has proved, as desire
Example Prove that, for all non-negative real numbers a, b,c, we have an inequality
a2 +bc
2a2+bc+
b2+ca
2b2+ca+
c2+ab
(7)Proof This inequality is equivalent to 2a2+ 2bc
2a2+bc +
2b2+ 2ca
2b2+ca +
2c2 + 2ab
2c2+ab ≥4
or
2a2+ 2bc
2a2+bc −1 +
2b2+ 2ca
2b2+ca −1 +
2c2+ 2ab
2c2+ab −1≥1
or
bc
2a2+bc+
ca
2b2+ca+
ab
2c2+ab ≥1
This is inequality (∗∗), so we are done
Example (Nguyen Dang Khoa) Given non-negative numbers a, b,c Show that
a2
8a2 + (b+c)2 +
b2
8b2+ (c+a)2 +
c2
8c2 + (a+b)2 ≤
1
Proof Applying AM - GM inequality and using inequality (∗∗), we have
a2
8a2+ (b+c)2 +
b2
8b2+ (c+a)2 +
c2
8c2 + (a+b)2 ≤
X a2
8a2+ 4bc =
1
X a2
2a2+bc ≤
1 So we are done
The equality occurs if and only if a =b =c or two numbers ofa, b,c are equal to
Example 10 (Vasile Cirtoaje) Given three non-negative real numbers satisfying
ab+bc+ca= Prove that
a2 + 1 +
1
b2+ 1 +
1
c2+ 1 ≥
3
Proof Since
a2+ 1 = 1−
a2
a2 + 1 the we can rewrite inequality in the form
a2
a2 + 1 +
b2
b2+ 1 +
c2
c2+ 1 ≤
3 or
a2
3a2+ 3 +
b2
3b2+ 3 +
c2
3c2+ 3 ≤
1 or
4a2
3a2 +ab+bc+ca +
4b2
3b2+ab+bc+ca +
4c2
3c2+ab+bc+ca ≤2
Using Cauchy - Schwarz inequality, we have 4a2
= (a+a)
2
≤ a
2
+ a
(8)Thus,
X 4a2
3a2+ab+bc+ca ≤
X a
a+b+c+
X a2
2a2 +bc ≤2
The equality holds if and only if a=b=cor a=b; c= and cyclic
Example 11 (Tigran Sloyan) Given a, b,c are positive real numbers Prove that
a2
(2a+b)(2a+c)+
b2
(2b+c)(2b+a)+
c2
(2c+a)(2c+b) ≤
Proof Applying Cauchy - Schwarz inequality, we obtain 9a2
(2a+b)(2a+c) =
(2a+a)2
2a(a+b+c) + (2a2+bc) ≤
2a a+b+c+
a2
2a2+bc
We write two similar inequalities and then add up all these relations We will find that
X 9a2
(2a+b)(2a+c) ≤
X 2a
a+b+c +
X a2
2a2+bc ≤3
Hence, we are done
Example 12 (Vo Quoc Ba Can) Given positive real numbers a, b, c Show that
a2
5a2 + (b+c)2 +
b2
5b2+ (a+c)2 +
c2
5c2 + (a+b)2 ≥
1
Proof UsingCauchy - Schwarz inequality, we get 9a2
5a2+ (b+c)2 =
(2a+a)2
(a2+b2+c2) + 2(2a2+bc) ≤
a2
a2+b2+c2 +
2a2
2a2+bc
Using that inequality and inequality (∗∗), we have
9X a
2
5a2+ (b+c)2 ≤
X a2
a2+b2 +c2 +
X a2
2a2+bc ≤3
Therefore, the initial inequality is proved
Example 13 (Nguyen Dang Khoa) Given non-negative real numbers a,b, csatisfying
a+b+c= Prove that
a2
4a2 + +bc+
b2
4b2+ +ca +
c2
4c2 + +ab ≤
(9)Proof The initial inequality is equivalent to 9a2
8a2+ + 2bc +
9b2
8b2+ + 2ca+
9c2
8c2+ + 2ab ≤
3 or
X 9a2
8a2+ (a+b+c)2+ 2bc ≤
3
Note that, we have (a+b+c)2 ≥max{4a(b+c),4b(c+a),4c(a+b)}
So we need to show that
X 9a2
8a2+ 4a(b+c) + 2bc ≤
3 Now, by using Cauchy - Schwarz, we get
9a2
8a2 + 4a(b+c) + 2bc =
(2a+a)2
4a(a+b+c) + 2(2a2+bc) ≤
a a+b+c+
a2
2(a2+ 2bc)
It implies
X 9a2
8a2+ + 2bc ≤
X a
a+b+c+
1
X a2
a2+ 2bc ≤
3 The equality holds if and only if b =c= 1, a= and cyclic
Example 14 Given a, b, c are positive real numbers have the sum is Prove that
ab
3ab+ 2b+c+
bc
3bc+ 2c+a +
ca
3ca+ 2a+b ≤
1
Proof Observe that
3ab+ 2b+c= 3ab+c(a+b+c) + 2b = (ab+bc+ca) + (2ab+c2) + 2b
UsingCauchy - Schwarz inequality, we have (1 + + 2)2
3ab+ 2b+c ≤
1
ab+bc+ca+
1 2ab+c2 +
2
b
Therefore,
16X ab
3ab+ 2b+c ≤
X
ab
ab+bc+ca+ ab
2ab+c2 + 2a
= +X ab
2ab+c2 ≤4
(10)2.2 Practice problems
Problem (Nguyen Dang Khoa) Given positive real numbers a, b, c such that abc = Prove that
(a+ 1)2
a+ +
(b+ 1)2
b+ +
(c+ 1)2
c+ ≤
3(a+b+c)
Problem Prove that, for all positive real numbers a, b, csuch that a+b+c= 1, we have
ab
3ab+ 2b+c+
bc
3bc+ 2c+a +
ca
3ca+ 2a+b ≤
1
Problem Given non-negative a,b, csatifying a+b+c= Show that
6a2+ 1 +
1 6b2+ 1 +
1 6c2+ 1 ≥
9
Problem Prove if a, b, c >0 and a+b+c= then
a a+bc+
b b+ca +
c c+ab ≥
3
Problem Given non-negative real number a,b, csuch that a+b+c >0 Prove that
a2
3a2 + (b+c)2 +
b2
3b2+ (c+a)2 +
c2
3c2 + (a+b)2 ≤
1
Problem Given a, b, c are positive real numbers Prove the following inequality
ab
a+ 3b+ 2c +
bc
b+ 3c+ 2a +
ca
c+ 3a+ 2b ≤
a+b+c
(11)3 Reference
[1] Vasile Cirtoaje, Algebraic inequality: Old and new methods, GIL Publishing house, 2004 [2] Vo Quoc Ba Can, Tran Quoc Anh, Using Cauchy - Schwarz method to prove inequalities, NXBDPSP
[3] Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, Inequalities with Beautiful Solutions, GIL Publishing house, 2009
[4] Pham Kim Hung, Secrets in Inequality, GIL Publishing house, 2007
[5] J.Michael Steele, The Cauchy - Schwarz Master Class, Cambridge University Press, 2004 [6] Mathematical Reflections