An Useful Technique in Proving Inequalities HSGS, Hanoi University of Science, Vietnam Abstract There are a lot of distinct ways to prove inequalities. This paper mentions a simple and useful technique in proving inequalities through problems. Basic knowledge A number α is called a root of the polynomial f(x) of multiplicity k (k 2, k N) if the following conditions are satisfied (i) f(x) is divisible by (x α) k, (ii) f(x) is not divisible by (x α) k+. In the other wds f(x) may be written in fm f(x) = (x α) k g(x), where g(x) is not divisible by x α. In the particular case if α is a root of f(x) of multiplicity 2, we say that f(x) has a double root x = α. We have the following result Theem. The necessary and sufficient condition so that α is a root of f(x) of multiplicity k as follows f (i) (α) = 0, i 0,,..., k }, f (k) (α) 0, where f (i) (x) = di f(x) (dx) i is the derivative of degree i of f(x) with convention f (0) (x) = f(x). 2 Examples Example. Prove that the following inequality holds f all positive real numbers a, b, c a a 2 + ab + 2b 2 + b b 2 + bc + 2c 2 + c c 2 + ca + 2a 2 a + b + c. 4 Analysis. Firstly we guess the equality holds when a = b = c. We want to find a result in fm as follows a a 2 ma + nb. + ab + 2b2 This inequality is equivalent to a (ma + nb)(a 2 + ab + 2b 2 ), ( m)a (m + n)a 2 b (2m + n)ab 2 2nb 0.
Consider the polynomial f(a) = ( m)a (m + n)a 2 b (2m + n)ab 2 2nb. We will find two real numbers m, n f which a = b is a double root of f(a). This happens when f(b) = 0 f (b) = 0 ( m)b (m + n)b (2m + n)b 2nb = 0 ( m)b 2 2(m + n)b 2 (2m + n)b 2 = 0 4m + 4n = 7m + n = Thus we give the solution as follows m = 9/6, n = 5/6. Solution. We will show that Indeed, this inequality is equivalent to which is clearly. Similarly, we have a 9a 5b a 2. + ab + 2b2 6 (a b) 2 (7a + 0b) 6(a 2 + ab + 2b 2 ) 0 b 9b 5c b 2, + bc + 2c2 6 Summing up these relations we obtain c 9c 5a c 2 + ca + 2a2 6. a a 2 + ab + 2b 2 + b b 2 + bc + 2c 2 + c 9a 5b 9b 5c 9c 5a c 2 + + + ca + 2a2 6 6 6 = a + b + c. 4 The proof is complete. Example 2. Let a, b, c be positive real numbers. Prove that 5a ab 2 a + b + 5b bc 2 b + c + 5c ca 2 c + a 2(a 2 + b 2 + c 2 ). Analysis. We observe that the equality occurs when a = b = c. Hence we want to find a result in fm as follows 5a ab 2 ma 2 + nb 2 a + b which is equivalent to (5 m)a ma 2 b ( + n)ab 2 nb 0. Consider the polynomial f(a) = (5 m)a ma 2 b (+n)ab 2 nb. Two real numbers m, n need to be chosen f which f(a) receives a = b as a double root. This happens if f(b) = f (b) = 0, i.e. m, n are roots of the system of equations (5 m) m ( + n) n = 0 (5 m) 2m ( + n) = 0 m =, n =. So we have the following solution 2
Solution. We will prove that Indeed, this result is equivalent to 5a ab 2 a + b a 2 b 2. (a b) 2 (2a + b) a + b which is obviously true. Similarly we also have 0 Therefe 5a ab 2 a + b 5b bc 2 b + c + 5b bc 2 b + c b 2 c 2, + 5c ca 2 c + a 5c ca 2 c + a c 2 a 2. (a 2 b 2 ) + (b 2 c 2 ) + (c 2 a 2 ) = 2(a 2 + b 2 + c 2 ) and we are done. Example. Let a, b, c be positive real numbers such that a + b + c = 9. Prove that a + b ab + 9 + b + c bc + 9 + c + a ca + 9 9. Analysis. We noticed that when a = b = c = the equality holds. Thus we need to find a result in fm a + b m(a + b) + n ab + 9 which is equivalent to a + b (ab + 9)(m(a + b) + n) 0. Consider the polynomial f(a) = a + b (ab + 9)(m(a + b) + n). We must find two real numbers m, n so that f(a) receives a = b = as a double root. This occurs if f(b) = f (b) = 0, where f (a) = a 2 b(m(a + b) + n) m(ab + 9). I.e. we have 2b (b 2 + 9)(2mb + n) = 0, b 2 b(2mb + n) m(b 2 + 9) = 0. From this system, choosing b = we get 54 8(6m + n) = 0, 27 (6m + n) 8m = 0, m =, n =. This analysis leads us to a solution below Solution. We will show that Indeed, we have Thus we only need to prove a + b ab + 9 a + b. () a + b ab + 9 4(a + b ) (a + b) (a + b) 2 + 6 (a + b) 2 + 6. (a + b) (a + b) 2 + 6 a + b,
(a + b) (a + b )((a + b) 2 + 6), (a + b 6) 2 0 which is trivially true. So () has been proved. Similar as above, we also have b + c bc + 9 b + c. (2) c + a ca + 9 c + a. () Summing up (), (2) and () side by side we get the desired inequality. Example 4 (Moldova, 2005). Let a, b, c be positive real numbers such that a 4 + b 4 + c 4 =. Prove that 4 ab + 4 bc + 4 ca. Analysis. We hope that there exists a result with the following fm 4 ab m(ab)2 + n, m(ab) 4m(ab) 2 + n(ab) 4n + 0. Consider the polynomial f(t) = mt 4mt 2 + nt 4n + (with t = ab). We want to find the numbers m, n so that f(t) receives t = as a double root. This happens if f() = f () = 0. That is m 4m + n 4n + = 0, m 8m + n = 0, m = /8, n = 5/8. Solution. We will show that 4 ab (ab)2 + 5. (4) 8 Indeed, this inequality may be rewritten as (ab) 4(ab) 2 + 5(ab) 2 0, (ab ) 2 (ab 2) 0. This result is true because = a 4 + b 4 + c 4 > a 4 + b 4 2(ab) 2 ab < Thus (4) has been proved. Similarly we also have 2 < 2. 4 bc (bc)2 + 5. (5) 8 4 ca (ca)2 + 5. (6) 8 4
Adding up the relations (4), (5) and (6) we obtain as desired. 4 ab + 4 bc + 4 ca (ab)2 + (bc) 2 + (ca) 2 + 5 8 a4 + b 4 + c 4 + 5 8 = Example 5. Let a, b, c be positive real numbers such that ab 2 + bc 2 + ca 2 =. Prove that 2a 5 + b 5 ab + 2b5 + c 5 bc + 2c5 + a 5 ca Analysis. By the given condition, we can guess that 5(a + b + c 2). This inequality is equivalent to 2a 5 + b 5 ab ma + nab 2 + pb. 2a 5 + b 5 ab(ma + nab 2 + pb ) 0. Consider the polynomial of degree 5 below f(a) = 2a 5 + b 5 ab(ma + nab 2 + pb ). We have f (a) = 0a 4 4ma b 2nab pb 4, f (a) = 40a 2ma 2 b 2nb, f (a) = 20a 2 24mab. We have to find the numbers m, n, p so that f(a) receives a = b as a root of multiplicity 4. This occurs when f(b) = f (b) = f (b) = f (b) = 0. That is 5b 5 b 2 (mb + nb + pb ) = 0, m + n + p = 5, 0b 4 4mb 4 2nb 4 pb 4 = 0, 4m + 2n + p = 0, m = 5, 40b 2mb 2nb n = 0, = 0, 2m + 2n = 40, 20b 2 24mb 2 = 0, p = 0. m = 5, This analysis leads us to the following solution Solution. We will show that 2a 5 + b 5 5a 0ab 2 + 0b. ab Indeed, this result is equivalent to (a b) 4 (2a + b) 0 which is clearly true. Similarly and 2b 5 + c 5 bc 2c 5 + a 5 ca 5b 0bc 2 + 0c, 5c 0ca 2 + 0a. 5
Adding up these relations we obtain 2a 5 + b 5 ab + 2b5 + c 5 bc + 2c5 + a 5 ca 5(a + b + c ) 0(ab 2 + bc 2 + ca 2 ) = 5(a + b + c 2). The proof is complete. Example 6. Prove that f all positive real numbers a, b, c 4a + 5b a 2 b + 0ab 2 + 4b + 5c b 2 c + 0bc 2 a + b b + c 5(a 2 + b 2 + c 2 ) (ab + bc + ca). + 4c + 5a c 2 a + 0ca 2 c + a Analysis. We believe that there exists a crect result in fm This is rewritten as 4a + 5b a 2 b + 0ab 2 a + b ma 2 + nb 2 ab f(a) = (4 m)a ma 2 b + ( n)ab 2 + (5 n)b 0. We want to choose m, n f which f(a) receives a = b as a double root. That is f(b) = f (b) = 0, m + n = 5, m + n = 2, m =, n = 4. Solution. We will check that Indeed, this inequality is equivalent to 4a + 5b a 2 b + 0ab 2 a + b a 2 + 4b 2 ab. a a 2 b ab 2 + b 0, (a b) 2 (a + b) 0 which is trivially true. Similarly 4b + 5c b 2 c + 0bc 2 b + c b 2 + 4c 2 bc, 4c + 5a c 2 a + 0ca 2 c + a Adding up these relations we get the required inequality. c 2 + 4a 2 ca. Example 7 (Crux). Let a, b, c be positive real numbers such that a 2 + b 2 + c 2 =. Find the minimum value of a E = b 2 + c 2 + b c 2 + a 2 + c a 2 + b 2. 6
Analysis. We guess that E minimizes at a = b = c = and we want to find a common fm of inequalities as follows which is rewritten as a b 2 + c 2 = a a 2 ma2 + n a ( a 2 )(ma 2 + n) f(a) = ma 4 (m n)a 2 + a n 0. The numbers m, n need to choose so that a = is a double root of f(a). Namely f( ) = f ( ) = 0, Thus we go to the following solution Solution. Firstly we will show that 2m + 6n =, 2m 6n =, m = 2, n = 0. Indeed this is equivalent to a a 2 2 a2. a a + 2 0 ( a ) 2 ( a + 2) 0 which is clearly true. Similarly b b 2 2 b2, Hence c c 2 2 c2. E 2 (a2 + b 2 + c 2 ) = 2. Thus we conclude that min E = 2, khi a = b = c =. Example 8 (Crux). Let a, b, c be positive real numbers such that Prove that a + + b + + c + = 2. 4a + + 4b + + 4c +. Analysis. We want to find two numbers m, n so that the following inequality holds 4a + m a + + n 4na 2 + (4m + 5n )a + m + n 0. 7
Let f(a) = 4na 2 + (4m + 5n )a + m + n. Note that if a = b = c = /2 the equality holds. So we must choose m, n so that a = /2 is a double root of f(a). This happens if f(/2) = f (/2) = 0. That is n + 4m+5n 2 + m + n = 0, 4n + 4m + 5n = 0, m =, n =. This analysis gives us a solution below Solution. We will check that Indeed, this inequality is equivalent to which is obviously true. Similarly Summing up these relations we get The proof is complete. 4a + a + (2a ) 2 0 4b + b +, 4c + c +. 4a + + 4b + + 4c + a + + b + + c + =. Remark. We have a me general result as follows: If a i > 0 (i =, 2,..., n) satisfy then a + + a 2 + + + a n + = n 4a + + 4a 2 + + + 4a n + 2n. Example 9 (Japan 997). Prove that the following inequality holds f all a, b, c > 0 (b + c a) 2 (c + a b)2 (b + c) 2 + + a2 (c + a) 2 + b 2 + (a + b)2 c 2 (a + b) 2 + c 2 5. Solution. Because the inequality is homogeneous, hence without loss of generality, we can assume that (nmalization) a + b + c =. Then the desired inequality reduces to ( 2a) 2 ( 2b)2 ( 2c)2 2a 2 + 6a + 9 2b 2 + 6b + 9 2c 2 6c + 9 5. We want to find an inequality having the following type ( 2a) 2 2a 2 ma + n, 6a + 9 f(a) = ( 2a) 2 (ma + n)(2a 2 6a + 9) 0. 8
The polynomial of degree three f(a) receives a = as a double root, the following conditions have to be satisfied f() = f () = 0. I.e. 5(m + n) = 0, m 2n + 4 = 0, m = 8/25, n = 2/25. So we will check that This result is equivalent to ( 2a) 2 2a 2 6a + 9 8a + 2. 25 (a ) 2 (2a + ) 0 which is trivially true. Similarly ( 2b) 2 2b 2 6b + 9 8b + 2, 25 These imply ( 2c) 2 2c 2 6c + 9 8c + 2. 25 and we are done. ( 2a) 2 ( 2b)2 ( 2c)2 8(a + b + c) + 69 2a 2 + 6a + 9 2b 2 + 6b + 9 2c 2 = 6c + 9 25 5 Example 0. Let a, b, c be non-negative real numbers such that a+b+c =. Find the maximum and minimum value of E = a 2 + a + 4 + b 2 + b + 4 + c 2 + c + 4. Analysis. We guess that the minimum value happens at the center, i.e. at a = b = c =. Therefe we will find two numbers m, n f which a 2 + a + 4 ma + n. We need the following system of equations (these are conditions to two graphs tangent each other) a 2 + a + 4 = ma + n, 2a+ 2 = m, a 2 +a+4 satisfied when a =. By this way we find m = 6 4, n = 6 4. Similar as above we also believe that the maximum value happens at the boundary, i.e. at (a, b, c) = (, 0, 0) and its permutions. Therefe, maybe the common inequality will be as a 2 + a + 4 αa + β where α, β are two numbers that we have to find so that the equality holds when a = and a = 0. By this way we obtain α = 2/ and β = 2. And we go to the following solution 9
Solution. (a) Find the minimum value: Firstly we will show that 6 a 2 + a + 4 4 a + 6 4. In deed, this result is equivalent to (a ) 2 0 which is obviously. Similarly 6 b 2 + b + 4 4 b + 6 4, 6 c 2 + c + 4 4 c + 6 4. Therefe E 6 Thus min E = 6, khi a = b = c =. (b) Find the maximum value: We will check that 4 9 6 (a + b + c) + = 6. 4 a 2 + a + 4 2 a + 2. This inequality is equivalent to a(a ) 0 which is true because 0 a. Similarly b 2 + b + 4 2 b + 2, c 2 + c + 4 2 c + 2. Thus E 2 (a + b + c) + 6 = 8. So max E = 8, when (a, b, c) = (, 0, 0) and its permutions. Example. Let a, b, c be positive real numbers such that a 2 + b 2 + c 2 =. Prove that 2 a + 2 b + 2 c. Solution. By the similar way as above, we need to check the following inequality 2 a a2 +. 2 But this result is equivalent to a(a ) 2 0, which is trivial. We also have similar relations Hence as desired. 2 b b2 +, 2 2 c c2 +. 2 2 a + 2 b + 2 c a2 + b 2 + c 2 + = 2 Example 2 (Mathematics and Youth magazine). Let a, b, c be positive real numbers such that a + b + c =. Prove that ( ) 0 ( + a 2 )( + b 2 )( + c 2 ). 9 0
Solution. The required inequality is equivalent to Now we find again two numbers α, β f which ln( + a 2 ) + ln( + b 2 ) + ln( + c 2 ) ln 0 9 ln( + a 2 ) αa + β. Note that the equality holds when a = b = c = /, hence we want two graphs of the functions f(a) = ln( + a 2 ) and g(a) = αa + β are tangent each other at point a = /. This means that f(/) = g(/) f (/) = g (/) ln 0 9 = α + β /5 = α α = /5, β = ln 0 9 5. So we will go to proving ln( + a 2 ) a 5 + ln 0 9 5. (7) Indeed, we consider the function h(a) = ln( + a 2 ) a 5 ln 0 9 + 5 2a +a 2 5 = a2 +0a 5(+a 2 ) with a [0, ]. We have h (a) =. The equation h (a) = 0 has a root a = / [0, ]. By establishing a table of increase and decrease of the function h(a) in the interval [0, ], we easily see that h(a) h(/) = 0. So we conclude that (7) is true. Similar as above we also have ln( + b 2 ) b 5 + ln 0 9 5 ln( + c 2 ) c 5 + ln 0 9 5 Adding up (7), (8) and (9) we get the desired result. The proof is complete. Remark 2. We have a me general result as follows: F x, x 2,..., x n 0 satisfy x + x 2 + + x n = then ( ( + x 2 )( + x 2 2) ( + x 2 n) + ) n n 2. Finally, we give a few problems f reader s practice (8) (9) Proposed problems. Let a, b, c, d be real numbers such that a 2 + b 2 + c 2 + d 2 = 4. Prove that a + b + c + d 8. 2. Let x, y, z be real numbers such that x + y + z =. Prove that + x 2 + + y 2 + + z 2 27 0.. (Poland 996) Let a, b, c 4 be real numbers such that a + b + c =. Prove that a a 2 + + b b 2 + + c c 2 + 9 0.
4. (USA 200) Prove that the following inequality holds f all positive real numbers a, b, c (2a + b + c) 2 (2b + c + a)2 (2c + a + b)2 2a 2 + + (b + c) 2 2b 2 + + (c + a) 2 2c 2 + (a + b) 2 8. 5. (China 2006) Let a, b, c be positive real numbers such that a + b + c =. Prove that a 2 + 9 2a 2 + (b + c) 2 + b 2 + 9 2b 2 + (c + a) 2 + c 2 + 9 2c 2 + (a + b) 2 5. 6. (France 2007) Let a, b, c, d be positive real numbers such that a + b + c + d =. Prove that 6(a + b + c + d ) a 2 + b 2 + c 2 + d 2 + 8. 7. Let a, b, c be positive real numbers such that a + b + c =. Prove that a 2 + b + c + b 2 + c + a + c 2 + a + b. 8. Let a, b, c be positive real numbers such that a 2 + b 2 + c 2 =. Find the minimum value of ( E = (a + b + c) + 2 a + b + ). c 9. Prove that f all positive real numbers a, b, c, a + 7b 2a + b + b + 7c 2b + c + c + 7a 2c + a + ab + bc + ca (a2 + b 2 + c 2 ). 0. (Crux) Let a, b, c be positive real numbers such that a 2 + b 2 + c 2 =. Prove that a + b + c + a + b + c 4.. (Mathematics and Youth magazine) Let x, y, z be positive real numbers such that x+2y + z = 4. Find the maximum value of E = 22y x 2xy + 24y 2 + 78z 8y 6yz + 54z 2 + 29x 27z zx + 6x 2. 2. (Crux) Let a, b, c > be real numbers such that Prove that a 2 + b 2 + c 2 =. a + + b + + c +.. Let a, b, c, d, e be positive real numbers such that Prove that 4 + a + 4 + b + 4 + c + 4 + d + 4 + e =. a 4 + a 2 + b 4 + b 2 + c 4 + c 2 + d 4 + d 2 + e 4 + e 2. 2
4. Prove that f all positive real numbers a, b, c a + b a + 2b + b + c b + 2c + c + a c + 2a 2 (a2 + b 2 + c 2 ). 5. Let a, b, c be non-negative real numbers such that a + b + c =. Prove that (a 2 + a + )(b 2 + b + )(c 2 + c + ) 27. 6. Let a, b, c be non-negative real numbers such that a + b + c =. Prove that a + 4 a 2 + 4 + b + 4 b 2 + 4 + c + 4 c 2 + 4. 7. Let a, b, c be non-negative real numbers such that a 2 + b 2 + c 2 = 2. Prove that References a + a 2 + 2 + b + b 2 + 2 + c + c 2 + 2 9 2. [] Mathematics and Youth Magazine, Education Publishing House of Vietnam. [2] Mathematical Reflections Journal. [] Titu Andreescu, Vasile Cirtoaje, Gabriel Dospinescu, Mircea Lascu, Old and New Inequalities, GIL Publishing House, 2004. [4] Dusan Djukic, Vladimir Jankovic, Ivan Matic, Nikola Petrovic, The IMO Compendium, Springer, 2004. Auth:, High School F Gifted Students, Hanoi University of Science, Vietnam. Email address: ngviethung0486@gmail.com