A New Note On Cauchy-Schwarz

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1 A New Note On Cauchy-Schwarz Hong Ge Chen January 30, student in Guangzhou university,guangzhou c- ity,china 1

2 As we can see,the Cauchy-Schwarz inequality is a very important inequality in proving inequalities no matter in mathematic contest or our daily use.there has many ways to apply the Cauchy-Schwarz method.now,i will introduce a nice way to use Cauchy-Schwarz that is very useful in prove three variables and fourth degrees inequality,also work on fifth degrees inequality. 1 A Simple Introduce Of pqr method In 2007,the inequality genius Võ Quôc Bá Cân published his article On a class of three-variable inequalities 1 in Mathematical Reflections.that gave the following conclusions. Let a, b, c are real numbers and set p = a + b + c, ab + bc + ca = p2 q 2 3, r = abcwe have a powerful estimate of r, Which is: (p + q) 2 (p 2q) r, (p ] q)2 (p + 2q) (1) it s a beautiful form of r. but,in fact,we usually set p = a + b + c, q = ab + bc + ca, r = abc that will change the form into: r 2p 3 + 9pq 2(p 2 3q) p 2 3q ], 2p 3 + 9pq + 2(p 2 3q) p 2 3q (2) I admit this is a ugly form of r and needs many computes.but it s obviously powerful. but,we can see that many lic inequality has something like a 2 b + b 2 c + c 2 a and a 3 b + b 3 c + c 3 a,that cause the inequality lic,without these things,the inequality should be a symmetric,a reason is: each three variables inequality can be rewrite as F (a, b, c)+g(a, b, c) (a k b m +b k c m +c k a m ) I wouldn t prove this fact here,you should remember all inequalities you have met before to see if it has this form. Now,Let us back to our topic,due to these reason.estimate y = a k b m +b k c m + c k a m become a important issue.but deal with unknown k, m is no double a difficult things,because different k, m get different ranges of y with different equality occurs.also with much complicated compute when k is a great value.here,we discuss about k = 2. When k = 2,we have to estimate a 2 b + b 2 c + c 2 a,we set p = a + b + c, q = ab + bc + ca, r = abc: also y 1 = a 2 b + b 2 c + c 2 a, y 2 = ab 2 + bc 2 + ca 2 1 1Vo Quoc Ba Can-On a class of three-variable inequalities,2007,mathematical Reflections ] 2

3 by Vieta Theorem { y1 + y 2 = pq 3r, y 1 y 2 = q 3 + p 3 r 6pqr + 9r 2 it s easy see that y 1, y 2 are the roots of following equation: When p, q are fixed.note: y 2 (pq 3r)y + q 3 + p 3 r 6pqr + 9r 2 = 0 (3) F (r, y) = y 2 (pq 3r)y + q 3 + p 3 r 6pqr + 9r 2 = 0 dy dr = F r(r, y) F y (r, y) = 3y + p3 6pqr + 9r 2 2y pq + 3r { 3y + p 3 6pq + 18r = 0 y 2 (pq 3r)y + q 3 + p 3 r 6pqr + 9r 2 = 0 Solve this equation,gives two roots: r 1 = 1 2p 3 + 9pq + (p 2 3q) ] p 2 3q y 1 = 1 9 p 3 2(p 2 3q) ] p 2 3q and r 2 = 1 y 2 = 1 9 2p 3 + 9pq (p 2 3q) ] p 2 3q p 3 + 2(p 2 3q) p 2 3q ] Having a simple look at r 1, r 2,we can see that both r 1, r 2 are in the range which we mentioned before.2 Hence the range of y is: p 3 2(p 2 3q) ] p 2 3q p 3 + 2(p 2 3q) ] p 2 3q, (4) 9 9 I believe you will feel very boring at solving equation which has such complicated coefficient.of course,the coefficient will be more ugly if the value of k, m is much large,although we can use Cauchy-Schwarz in the estimation(see kuing s new article).if you don t think so,please try k = 3, m = 1 I will introduce a simple way in next section. 3

4 2 Choose a suitable Polynomial to applying Cauchy- Schwarz Why we have to estimate so many times? as you can see,estimate a 2 b + b 2 c + c 2 a and a 3 b + b 3 c + c 3 a is not a easy job,the question come in natural.can we find some general polynomial that can estimate some value of k and m in once time? Let s analysis this problem in detail.the reasone we need to estimate for so many times is that: Although we have: a 3 b + b 3 c + c 3 a = (a 2 b + b 2 c + c 2 a)(a + b + c) abc(a + b + c) a 2 b 2 We still can t use the range of (a 2 b+b 2 c+c 2 a) to get the range of a 3 b+b 3 c+c 3 a. Why? The reasone is: if p, q are fixed.we can t sure the (a 2 b + b 2 c + c 2 a) and abc(a + b + c) a 2 b 2 = rp q 2 will obtain the max or min value at the same time in the define range of r. I hope I have explained clearly,let s give a simple example to this principle: Example: set f(x), g(x) C1, 4], f(x) obtain its min value when x = 1 and max value when x = 4,g(x) obtain its min value in x = 2 and max value in x = 3,Note H(x) = f(x) + g(x).do you know when H(x) obtain the max and min value? certainly not! it s unknown. Now,we find a regular pattern,if the rest things don t have r,only have p, q then we can use the result we found before,therefore,choose the following polynomial is suitable. S k = a 2 b + b 2 c + c 2 a + kabc (5) Here we will use another way to estimate this polynomial. f(a, b, c) = a 2 b + b 2 c + c 2 a + kabc + λ 1 (a + b + c) + λ 2 (ab + bc + ca) by Lagrange multiplier method: 2ab + c 2 + kbc + λ 1 + λ 2 (b + c) = 0 2bc + a + kca + λ 1 + λ 2 (a + c) = 0 2ca + b 2 + kab + λ 1 + λ 2 (b + a) = 0 a + b + c ab + bc + ca = p = q (a + b + c) 2 + k(ab + bc + ca) + 3λ 1 + 2λ 2 (a + b + c) = 0 4

5 2a b c 6ab + 3c 2 + 3kbc p 2 kq = 3λ 1 = (p 2 + kq + 2pλ 2 ) 2b a c 6bc + 3a 2 + 3kac p 2 kq = this system which guide us to use Cauchy-Schwarz : 2c b a 6ca + 3b 2 + 3kab p 2 kq 2 ] (2a b c)(6ab + 3c 2 + 3kbc p 2 kq)] (2a b c) ] 2 (6ab + 3c 2 + 3kbc p 2 kq) 2 there are some identities available: ( (2a b c) 2 = 6 a 2 ) ab (6) (6ab + 3c 2 + 3kbc p 2 kq) 2 = 9 a k ab 3 + (9k 2 + 3k) a 2 b a 2 bc (6p 2 + 6kq) a 2 (12kq + 6kp 2 + 6k 2 q + 12p 2 ) ab + 3p 4 + 3k 2 q 2 + 6kq 2 The second one is a little bit ugly.but if we transform in to p, q, r,it s much simpler and beautiful. (6ab + 3c 2 + 3kbc p 2 kq) 2 = (12k 36)p 2 q + (6k k)q 2 + 6p 4 18pS k k (8) (2a b c) 2 = 6( a 2 ab) = 6(p 2 2q) (9) (2a b c)(6ab + 3c 2 + 3kbc p 2 kq) = S k (9 + 3k)pq 3p(p 2 3q) Therefore, Sk (9 + 3k)pq 3p(p 2 3q) ] 2 6(p 2 2q) (12k 36)p 2 q + (6k k)q 2 + 6p 4 18pkS k ] it gives the range of S k is: (3 2k)p 3 + 9kpq 2 ] (k 2 3k + 9)(p 2 3q) 3, (7) (10) (3 2k)p 3 + 9kpq + 2 ] (k 2 3k + 9)(p 2 3q) 3 5

6 3 Application In the last section,we have gotten the range of S k is (3 2k)p 3 + 9kpq 2 ] (k 2 3k + 9)(p 2 3q) (3 3 2k)p 3 + 9kpq + 2 ] (k 2 3k + 9)(p 2 3q) 3 Now,Let s see the application of this powerful tool.in order to be more convince,let s build a lemma at first. lemma: a 3 b + b 3 c + c 3 a p4 + 9p 2 q q 2 2(p 2 3q) 7p 2 (p 2 3q) and a 3 b + b 3 c + c 3 a p4 + 9p 2 q q 2 + 2(p 2 3q) 7p 2 (p 2 3q) Võ Quôc Bá Cân called this as pqr lemma,where p = a+b+c, q = ab + bc + ca Proof of the lemma: Just notice that: (a 3 b + b 3 c + c 3 a) = (a 2 b + b 2 c + c 2 a + abc)(a + b + c) (ab + bc + ca) 2 = S 1 p q 2 and using the range of S k for k = 1,The result follows immediately. Example 1 Let a, b, c R,Prove that: 1 a 3 b + b 3 c + c 3 a 7 3 (a 2 + b 2 + c 2 ) 2 8, (Vasile Cirtoaje,Dan Chen) For the left side of this inequality,it s really famous since Vasile Cirtoaje found it in 1992,there also has many nice proofs 2 of it.also we can use Vo Quoc Ba Can s sum of square technique 3 to solve it.but now,i will use the theory I talked before to prove this problem. proof:(by kuing) For the left side (a 2 + b 2 + c 2 ) 2 3(a 3 b + b 3 c + c 3 a) 2 Prove Vasc s inequality by Cauchy-Schwarz 3 The Sum of Square technique,vo Quoc Ba Can-Pham Thi Hung 6

7 Using the lemma,it s suffice to check Or (p 2 2q) 2 3 p4 + 9p 2 q q p 2 (p 2 3q) 3 9(2p 2 7q) 2 (p 2 3q) 2 0 Which is obviously true.equality occurs when p 2 = 3q Or 2p 2 = 7q.For the right side.using the lemma again,it s enough to prove that: 8 p4 + 9p 2 q q 2 2 7p 2 (p 2 3q) 3 and after expand gives 7(p 2 qa) 2 (8 + 7)p 4 + ( )p 2 q + ( )q p 2 (p 2 3q) 3 Due to p 2 3q,we can get that: (8 + 7)p 4 + ( )p 2 q + ( )q 2 ] p 2 (p 2 3q) 3 factor it,gives (2+ 7)p 2 +2q] 2 ( )p 4 +( )p 2 q +( )q 2 ] 0 Or ( )p 4 + ( )p 2 q + ( )q 2 0 actually,this inequality is hold in the condition p 2 3q Hence we are done! Example 2 Let x, y, z are real numbers,prove that: x 4 + y 4 + z 4 + 2xyz(x + y + z) x 3 y + y 3 z + z 3 x proof:as we can see,the inequality can be strength as (Vasile Cirtoaje) x 4 + 2xyz(x + y + z) (x 3 y + y 3 z + z 3 x) 2 (xy + yz + xz)2 3 Or 3( x 2 ) 2 8( xy) 2 3 x 3 y 6xyz( x)] 0 Normalize p = 3 3(p 2 2q) 2 8q 2 + 3(S 5 p q 2 ) 9S 5 7q 2 108q

8 Using our theory,we have: S q (3 q) 3 9 so it s enough to prove that: q 2 9q (3 q) 3 Or Which is prefectly true! Hence we are done! q 2 (q 3) 2 0 8

On a class of three-variable inequalities. Vo Quoc Ba Can

On a class of three-variable inequalities Vo Quoc Ba Can 1 Theem Let a, b, c be real numbers satisfying a + b + c = 1 By the AM - GM inequality, we have ab + bc + ca 1, therefe setting ab + bc + ca = 1