Shi and Wu Journal of Inequalities and Applications 2013, 2013:329 http://www.journalofinequalitiesandapplications.com/content/2013/1/329 RESEARCH Open Access An artificial proof of a geometric inequality in a triangle Shi-Chang Shi1* and Yu-Dong Wu2 * Correspondence: 532686108@qq.com Department of Education, Zhejiang Teaching and Research Institute, Hangzhou, Zhejiang 310012, People’s Republic of China Full list of author information is available at the end of the article Abstract In this paper, the authors give an artificial proof of a geometric inequality relating to the medians and the exradius in a triangle by making use of certain analytical techniques for systems of nonlinear algebraic equations MSC: 51M16; 52A40 Keywords: geometric inequality; triangle; medians; inradius; circumradius Introduction and main results For a given ABC, let a, b and c denote the side-lengths facing the angles A, B and C, respectively Also, let ma , mb and mc denote the corresponding medians, , rb and rc the corresponding exradii, s = (a + b + c) the semi-perimeter, the area In addition, we let m = (b + c) – a = m = a + (b + c) , s(s – a), and r = √ a s(s – a) (s – a) Throughout this paper, we will customarily use the cyclic sum symbols as follows: f (a) = f (a) + f (b) + f (c) and f (b, c) = f (a, b) + f (b, c) + f (c, a) In , Liu [] found the following interesting geometric inequality relating to the medians and the exradius in a triangle with the computer software BOTTEMA invented by Yang [–], and Liu thought this inequality cannot be proved by a human Theorem . In ABC, the best constant k for the following inequality (rb – rc ) ≥ k · (mb – mc ) (.) © 2013 Shi and Wu; licensee Springer This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited Shi and Wu Journal of Inequalities and Applications 2013, 2013:329 http://www.journalofinequalitiesandapplications.com/content/2013/1/329 Page of 13 is the real root on the interval (, ) of the following equation ,k – ,k – ,k – ,k + , = (.) Furthermore, the constant k has its numerical approximation given by . In this paper, the authors give an artificial proof of Theorem . Preliminary results In order to prove Theorem ., we require the following results Lemma . In ABC, if a ≤ b ≤ c, then ra + rb + rc – r + m ≥ s(s – a)(b – c) (s – b)(s – c) Proof From a = (s – b) + (s – c) and the formulas of the exradius = etc., we get (.) s–a = √ s(s–a)(s–b)(s–c) , s–a ra + rb + rc – r + m = a s(s – a) s(s – a)(s – b)(s – c) – + + – s(s – a) (s – a) (s – b) (s – c) (s – a) = a (s – b)(s – c) (s – b)(s – c) (s – b)(s – c) s(s – a) + + – – (s – a) (s – b) (s – c) (s – a) = (s – b)(s – c) – a s–c s–b s(s – a) + – + (s – a) s–b s–c = (b – c) (b – c) s(s – a) – + (s – a) (s – b)(s – c) = s(s – a)(b – c) – (s – b)(s – c) (s – a) (.) For a ≤ b ≤ c, we have s – a ≥ s – b ≥ s – c > , then < ≤ ≤ , s–a s–b s–c hence ≥ > (s – b)(s – c) (s – a) Inequality (.) follows from inequalities (.)-(.) immediately (.) Shi and Wu Journal of Inequalities and Applications 2013, 2013:329 http://www.journalofinequalitiesandapplications.com/content/2013/1/329 Lemma . In Page of 13 ABC, we have (mb + m )(mc + m ) ≥ s (s – b)(s – c) (.) and a(mb + mc ) – s(s – b)(s – c) ≥ √ s (s – b)(s – c)(b – c) a (.) Proof of inequality (.) From b+c a– m – as = ≥ , we immediately obtain m ≥ as (.) In view of the AM-GM inequality, we get a (s – b) + (s – c) = ≥ (s – b)(s – c) (.) By the power mean inequality, we have a = (s – b) + (s – c) ≥ √ s–b+ By the well-known inequalities mb ≥ ities (.)-(.), we obtain √ √ s–c s(s – b) and mc ≥ (mb + m )(mc + m ) ≥ =s s(s – b) + √ s–b+ as a s(s – c) + √ s–c+ as a √ √ a( s – b + s – c) + =s a+ ≥s √ √ ( s – b + s – c) + (s – b)(s – c) =s a + (s – b)(s – c) (s – b)(s – c) ≥ s (s – b)(s – c) The proof of inequality (.) is thus complete (.) √ s(s – c), together with inequal- Shi and Wu Journal of Inequalities and Applications 2013, 2013:329 http://www.journalofinequalitiesandapplications.com/content/2013/1/329 Page of 13 Proof of inequality (.) According to the well-known inequalities mb ≥ √ s(s – c) and inequality (.), we have √ s(s – b), mc ≥ a(mb + mc ) – s(s – b)(s – c) = a – (s – b)(s – c) (mb + mc ) + (s – b)(s – c) (mb + mc ) – s (s – b)(s – c) ≥ a – (s – b)(s – c) · mb mc + (s – b)(s – c) s(s – b) + s(s – c) – s (s – b)(s – c) ≥ s a – (s – b)(s – c) (s – b)(s – c) + (s – b)(s – c) a – (s – b)(s – c) = s (s – b)(s – c) a – (s – b)(s – c) √ s (s – b)(s – c)(b – c) = √ a + (s – b)(s – c) √ s (s – b)(s – c)(b – c) ≥ a (.) Hence, we complete the proof of inequality (.) Lemma . In ABC, we have mb mc ≤ m (.) Proof From the formulas of the medians, we have mb mc – m = = = mb mc – m mb mc + m (c + a – b )(a + b – c ) – (a + (b + c) ) mb mc + m {[a – (b + c) ] – (b + c + bc)}(b – c) ≤ (mb mc + m ) Therefore, inequality (.) holds true Lemma . In ABC, if a ≤ b ≤ c, then m m mb + m c + + ma + m m + mb m + mc ≥ – (b + c) (mb + mc ) m m a(b + c) + – m m s(s – b)(s – c) Proof It is obvious that mb > c – (mb – mc ) = b (.) and mc > b – c , then we have mb + mc > (b + c), thus (mb – mc ) (b + c) (b – c) = ≤ (b – c) (mb + mc ) (mb + mc ) (.) Shi and Wu Journal of Inequalities and Applications 2013, 2013:329 http://www.journalofinequalitiesandapplications.com/content/2013/1/329 Page of 13 For a ≤ b ≤ c, we have that ⎧ ⎨m ma ≥ ⎩mb ≥ m ≥ mc (.) By Lemma . and inequalities (.)-(.), we have m b + mc m m + + ma + m m + mb m + mc – m m – m m = mb + mc – m m (m – ma ) m (m – mb mc ) + + ma + m m (ma + m ) m (m + mb )(m + mc ) ≥ m (m – ma ) (mb + mc ) – m + (ma + m )(mb + mc + m ) m (ma + m ) = (mb + mc ) – (mb – mc ) – m m (m – ma ) + (ma + m )(mb + mc + m ) m (ma + m ) = (b – c) – (mb – mc ) m (b – c) – (ma + m )(mb + mc + m ) m (ma + m ) ≥ (b – c) – (b – c) (b – c) – (ma + m )(mb + mc + m ) (ma + m ) = (b – c) –(b – c) – (ma + m )(mb + mc + m ) (ma + m ) ≥ (b – c) –(b – c) – (ma + m )(mb + mc ) (ma + m ) = –(b – c) (b – c) – (mb + mc ) (mb + mc ) ≥ –(b – c) (mb + mc ) (.) By inequality (.), (.) and a ≤ b ≤ c, we obtain that (b + c) a(b + c) – s(s – b)(s – c) (mb + mc ) (b + c) [a(mb + mc ) – s(s – b)(s – c)] s(s – b)(s – c)(mb + mc ) √ (b + c) s (s – b)(s – c)(b – c) ≥ · s(s – b)(s – c)(mb + mc ) a = = (b + c) (b – c) √ a (s – b)(s – c)(mb + mc ) ≥ (b + c) (b – c) a (mb + mc ) ≥ (b – c) (mb + mc ) (.) Shi and Wu Journal of Inequalities and Applications 2013, 2013:329 http://www.journalofinequalitiesandapplications.com/content/2013/1/329 Page of 13 By inequalities (.)-(.), we have mb + mc m m + + ma + m m + mb m + mc – = – (b + c) (mb + mc ) m a(b + c) m + – m m s(s – b)(s – c) m b + mc m m + + ma + m m + mb m + mc + – m m – m m (b + c) a(b + c) – s(s – b)(s – c) (mb + mc ) ≥ (b – c) –(b – c) + (mb + mc ) (mb + mc ) = (b – c) ≥ (mb + mc ) (.) Inequality (.) follows from inequality (.) immediately Lemma . In ABC, if a ≤ b ≤ c, then m (b + c) m + + ≥ m m a (.) and √ m + a √ ≤ s (.) Proof Without loss of generality, we can take b + c = and a = x, for a ≤ b ≤ c, we have < x ≤ (i) First, we prove inequality (.) m (b + c) m + + –= m m a = ≥ + x + – x – x + – + x x + x ( – x )( + x ) + x · (–x )+(+x ) + + ( – x ) – x ( – x ) – x = ( – x ) + x + – ( + x ) x = (x – ) ( – x ) + ( + x ) x ≥ (x – ) ( – x ) + = ( – x ) ≥ Inequality (.) terminates the proof of inequality (.) (.) Shi and Wu Journal of Inequalities and Applications 2013, 2013:329 http://www.journalofinequalitiesandapplications.com/content/2013/1/329 Page of 13 (ii) Second, we prove inequality (.) m + √ √ a – s √ √ = ( – x) –x – √ √ = – x + x – ( – x) √ – – x( – x) ≤ =√ √ + x + ( – x) (.) Inequality (.) follows from inequality (.) immediately Lemma . In ABC, if a ≤ b ≤ c, then s(s – a)(b – c) ma mb + mb mc + mc ma – m m – m ≥ (b – c) – (s – b)(s – c) Proof By the AM-GM inequality, the well-known inequalities mb ≥ √ s(s – c), we get √ (.) s(s – b) and mc ≥ (mb + mc ) ≥ mb mc ≥ s (s – b)(s – c) ≥ a (s – b)(s – c) ≥ (s – b)(s – c) or √ mb + mc ≥ (s – b)(s – c) (.) By inequalities (.), (.), (.), (.), (.), we obtain that ma mb + mb mc + mc ma – m m – m = (mb + mc )(ma – m ) m (mb – m ) m (mc – m ) (mb – mc ) + + – + (b – c) ma + m mb + m mc + m (mb + mc ) = (mb + mc )(b – c) m (b + c)(c – b) m (b + c)(b – c) + + (ma + m ) (mb + m ) (mc + m ) – = (b + c) (b – c) + (b – c) (mb + mc ) m (b – c) m (b – c) (mb + mc )(b – c) + + (ma + m ) (mb + m ) (mc + m ) – m (b + c) (b – c) (mb + m )(mc + m )(mb + mc ) – (b + c) (b – c) + (b – c) (mb + mc ) m (b + c) m m a(b + c) + – √ (b – c) + – m m s(s – b)(s – c) s(s – b)(s – c) √ m m (m + a)(b + c) + = + – (b – c) √ m m s(s – b)(s – c) ≥ Shi and Wu Journal of Inequalities and Applications 2013, 2013:329 http://www.journalofinequalitiesandapplications.com/content/2013/1/329 Page of 13 ≥ m m (b + c) + (b – c) + – m m (s – b)(s – c) = m m [(b + c) – a ] [(b + c) – a ] + + (b – c) + – m m (s – b)(s – c) (s – b)(s – c) ≥ [(b + c) – a ] m m s(s – a) + (b – c) + – + m m (s – b)(s – c) a = m m (b + c) s(s – a) – (b – c) + + – m m a (s – b)(s – c) ≥ s(s – a) – – (b – c) (s – b)(s – c) s(s – a)(b – c) = (b – c) – (s – b)(s – c) The proof of Lemma . is thus completed Lemma . In ABC, if inequality (.) holds, then k ≤ Proof Let b = c = and a = x For a ≤ b ≤ c, we have x ∈ (, ], then inequality (.) is equivalent to √ √ – x x – x – ( – x) ⇐⇒ ⇐⇒ √ ≥ k – x – √ x + ( + x) +x ≥k· √ √ –x ( – x + x + ) √ √ ( + x)( – x + x + ) k≤ ( – x)( + x) (.) Taking x = in inequality (.), we obtain that k ≤ Lemma . In ABC, if a ≤ b ≤ c and < k ≤ , then we have (rb – rc ) – k · (mb – mc ) ≥ (r – m ) – k(m – m ) Proof For (rb – rc ) = ra – rb rc = ra – s and (mb – mc ) = ma – m b mc = a – mb mc , hence, by Lemmas . and ., we have (rb – rc ) – k · = (mb – mc ) – (r – m ) + k(m – m ) ra – r – m + k mb mc – m m – m – (b – c) (.) Shi and Wu Journal of Inequalities and Applications 2013, 2013:329 http://www.journalofinequalitiesandapplications.com/content/2013/1/329 ≥ s(s – a)(b – c) ks(s – a)(b – c) – (s – b)(s – c) (s – b)(s – c) = ( – k)s(s – a)(b – c) ≥ (s – b)(s – c) Page of 13 The proof of Lemma . is complete Lemma . (see [, , ]) Define F(x) = a xn + a xn– + · · · + an , and G(x) = b xm + b xm– + · · · + bm If a = or b = , then the polynomials F(x) and G(x) have a common root if and only if a R(F, G) := b a a b b ··· a an ··· ··· b a bm ··· b ⎫ ⎪ ⎪ ⎪ ⎬ an a ⎪ ⎪ ⎪ ⎭ ··· an b = , ⎫ ⎪ ⎪ ⎪ ⎬ bm m ··· bm ⎪ ⎪ ⎪ ⎭ n where R(F, G) ((m + n) × (m + n) determinant) is Sylvester’s resultant of F(x) and G(x) Lemma . (see [, ]) Given a polynomial f (x) with real coefficients f (x) = a xn + a xn– + · · · + an , if the number of the sign changes in the revised sign list of its discriminant sequence D (f ), D (f ), , Dn (f ) is v, then the number of the pairs of distinct conjugate imaginary roots of f (x) equals v Furthermore, if the number of non-vanishing members in the revised sign list is l, then the number of the distinct real roots of f (x) equals l – v The proof of Theorem 1.1 Proof If k ≤ ,we can easily find that inequality (.) holds Hence, we only need to consider the case k > , and by Lemma ., we only need to consider the case < k ≤ Now we determine the best constant k such that inequality (.) holds Since inequality (.) is symmetrical with respect to the side-lengths a, b and c, there is no harm in supposing a ≤ b ≤ c Thus, by Lemma ., we only need to determine the best constant k such Shi and Wu Journal of Inequalities and Applications 2013, 2013:329 http://www.journalofinequalitiesandapplications.com/content/2013/1/329 Page 10 of 13 that (r – m ) – k(m – m ) ≥ or, equivalently, that a (b + c) – a – (b + c – a) (b + c) – a –k (b + c) – a a + (b + c) – ≥ (.) Without loss of generality, we can assume that a=t and b+c = ( < t ≤ ), because inequality (.) is homogeneous with respect to a and (.) is equivalent to the following inequality: √ √ t – t – t – ( – t) √ –k – t – √ t + b+c Thus, clearly, inequality ≥ (.) We consider the following two cases separately Case When t = , inequality (.) holds true for any k ∈ R := (–∞, +∞) Case When < t < , inequality (.) is equivalent to the following inequality: √ √ ( + t)( – t + t + ) k≤ ( – t)( + t) (.) Define the function √ √ ( + t)( – t + t + ) g(t) := , ( – t)( + t) x ∈ (, ) Calculating the derivative for g(t), we get √ √ √ √ √ ( – t + t + ) · – t · [(t + t + t – ) – ( – t) – t · t + ] g (t) = √ √ ( – t) ( + t) t + · – t By setting g (t) = , we obtain √ √ √ – t · t + t + t – – ( – t) – t · t + = (.) √ It is easily observed that the equation – t = has no real root on the interval (, ) Hence, the roots of equation (.) are also solutions of the following equation: √ √ t + t + t – – ( – t) – t · t + = , that is, ( + t) ϕ(t) = , (.) Shi and Wu Journal of Inequalities and Applications 2013, 2013:329 http://www.journalofinequalitiesandapplications.com/content/2013/1/329 Page 11 of 13 where ϕ(t) = t + t – It is obvious that the equation ( + t) = (.) has no real root on the interval (, ) It is easy to find that the equation ϕ(t) = (.) has one positive real root Moreover, it is not difficult to observe that ϕ() = – < and ϕ() = > We can thus find that equation (.) has one distinct real root on the interval (, ) So that equation (.) has only one real root t given by t = . on the interval (, ), and g(t)max = g(t ) ≈ . ∈ (, ) (.) Now we prove g(t ) is the root of equation (.) For this purpose, we consider the following nonlinear algebraic equation system: ⎧ ⎪ ϕ(t ) = , ⎪ ⎪ ⎪ ⎪ ⎨t + – u = , ⎪ ⎪ – t – v = , ⎪ ⎪ ⎪ ⎩ ( + t)(u + v ) – ( – t)( + t) k = (.) It is easy to see that g(t ) is also the solution of nonlinear algebraic equation system (.) If we eliminate the v , u and t ordinal by the resultant (by using Lemma .), then we get ,,,,, · φ (k) · φ (k) = , (.) where φ (k) = ,k – ,k – ,k – ,k + , and φ (k) = k – ,k + ,k – ,k + , The revised sign list of the discriminant sequence of φ (k) is given by [, , –, –] (.) Shi and Wu Journal of Inequalities and Applications 2013, 2013:329 http://www.journalofinequalitiesandapplications.com/content/2013/1/329 Page 12 of 13 The revised sign list of the discriminant sequence of φ (k) is given by [, , –, –] (.) So the number of sign changes in the revised sign list of (.) and (.) are both Thus, by applying Lemma ., we find that the equations φ (k) = (.) φ (k) = (.) and both have two distinct real roots In addition, it is easy to find that φ () = , > ; φ () = , > , φ () = –, < ; φ () = –, < , φ () = –, < ; φ () = –, < φ () = , > ; φ () = , > and We can thus find that equation (.) has two distinct real roots on the intervals (, ) and (, ) And equation (.) has two distinct real roots on the intervals (, ) and (, ) Hence, by (.), we can conclude that g(t ) is the root of equation (.) The proof of Theorem . is thus completed Competing interests The authors declare that they have no competing interests Authors’ contributions Both authors contributed equally and read and approved the final manuscript Author details Department of Education, Zhejiang Teaching and Research Institute, Hangzhou, Zhejiang 310012, People’s Republic of China Department of Mathematics, Zhejiang Xinchang High School, Shaoxing, Zhejiang 312500, People’s Republic of China Acknowledgements The authors would like to thank the anonymous referees for their very careful reading and making some valuable comments which have essentially improved the presentation of this paper Received: 28 January 2013 Accepted: July 2013 Published: 19 July 2013 Shi and Wu Journal of Inequalities and Applications 2013, 2013:329 http://www.journalofinequalitiesandapplications.com/content/2013/1/329 References Liu, B-Q: BOTTEMA, What We Have Seen - the New Theory, New Method and New Result of the Research on Triangle Tibet People Press, Lhasa (2003) (in Chinese) Yang, L: A dimension-decreasing algorithm with generic program for automated inequality proving High Technol Lett 25(7), 20-25 (1998) (in Chinese) Yang, L: Recent advances in automated theorem proving on inequalities J Comput Sci Technol 14(5), 434-446 (1999) Yang, L, Xia, B-C: Automated Proving and Discovering on Inequalities Science Press, Beijing (2008) (in Chinese) Yang, L, Xia, S-H: Automated proving for a class of constructive geometric inequalities Chinese J Comput 26(7), 769-778 (2003) (in Chinese) Sylvester, JJ: A method of determining by mere inspection the derivatives from two equations of any degree Philos Mag 16, 132-135 (1840) Yang, L, Zhang, J-Z, Hou, X-R: Nonlinear Algebraic Equation System and Automated Theorem Proving, pp 23-25 Shanghai Scientific and Technological Education Press, Shanghai (1996) (in Chinese) Yang, L, Hou, X-R, Zeng, Z-B: A complete discrimination system for polynomials Sci China Ser E 39(7), 628-646 (1996) doi:10.1186/1029-242X-2013-329 Cite this article as: Shi and Wu: An artificial proof of a geometric inequality in a triangle Journal of Inequalities and Applications 2013 2013:329 Page 13 of 13 ... doi:10.1186/1029-242X-2013-329 Cite this article as: Shi and Wu: An artificial proof of a geometric inequality in a triangle Journal of Inequalities and Applications 2013 2013:329 Page 13 of 13 ... contributions Both authors contributed equally and read and approved the final manuscript Author details Department of Education, Zhejiang Teaching and Research Institute, Hangzhou, Zhejiang 310012,... ) ≥ Inequality (.) terminates the proof of inequality (.) (.) Shi and Wu Journal of Inequalities and Applications 2013, 2013:329 http://www.journalofinequalitiesandapplications.com/content/2013/1/329