Seven geometric inequalites proposed by tran quang hung

Trn Quang Hng - High school for gifted students at science Some geometric inequalities Tran Quang Hung Problem With arbitrary triangle ABC inscribed (O), incenter I and an arbitrary point M in small arc BC Prove that MA + 2OI ≥ MB + MC ≥ MA − 2OI Proof By the projection of vectors we have −−→ −−→ −−→ −−→ MA MA = 2MO.MA ⇒ MA = 2MO MA similar −−→ −−→ −−→ MB −−→ MC MB = 2MO , MC = 2MO MB MC From them we have −−→ −−→ −−→ −−→ MB MC MA MB + MC − MA = 2MO.( + − ) (1) MB MC MA By Cauchy-Swart inequality we have −−→ −−→ −−→ −−→ −−→ −−→ −−→ −−→ −−→ −−→ MB MC MA MB MC MA MB MC MA + − | ≤ MO · ( + − ) ≤ MO| + − | (2) −MO| MB MC MA MB MC MA MB MC MA −−→ −−→ −−→ MB MC MA We have MO = R, we will calculate | + − | MB MC MA −−→ −−→ −−→ MB MC MA + − | | MB MC MA −−→ −−→ −−→ −−→ −−→ −−→ MB MA MC MA MB MC − − ) = + 2( MB MC MB MA MC MA −−→ −−→ −−→ −−→ −−→ −−→ = + 2(cos(MB, MC) − cos(MB, MA) − cos(MC, MA)) = − 2(cos A + cos B + cos C) (Because M in small arc BC) R+r R+r = 3−2 (Here we use the well-know equality: cos A + cos B + cos C = ) R R R2 − 2Rr = R2 OI = (Here we use the Euler’s formula) R From this we have −−→ −−→ −−→ MB MC MA MO| + − | = OI (3) MB MC MA Thus from (1), (2), (3) we have the inequality MA + 2OI ≥ MB + MC ≥ MA − 2OI and equality when ABC is equaliteral triangle We completed the solution Trn Quang Hng - High school for gifted students at science Problem Given are value of sum ABC orthorcenter H, circumradius R, with any M on plane find minimum MA3 + MB + MC − R · MH 2 Proof By AM-GM inequality we have MA3 R2 + MA2 MA3 + ≥ + R.MA ≥ 2MA2 R R ⇒ Similar we have R2 MA3 ≥ MA2 − R 2 MB 3 R2 MC 3 R2 ≥ MB − , ≥ MC − R 2 R 2 Thus MA3 + MB + MC 3 ≥ (MA2 + MB + MC ) − R2 R 2 Called O is circumcenter of ABC (1) MA2 + MB + MC −−→ −→ −−→ −−→ −−→ −→ = (MO + OA)2 + (MO + OB)2 + (MO + OC)2 −−→ −→ −−→ −→ = 3MO + 2MO(OA + OB + OC) + 3R2 −−→ −−→ = 3MO + 2MO.OH + 3R2 = 3MO − (OM + OH − MH ) + 3R2 = 2MO − OH + MH + 3R2 ≥ 3R2 − OH + MH (2) −→ −−→ −→ −−→ (Here we use familiar equal of vector OA + OB + OC = OH) Form (1), (2) we have 3 MA3 + MB + MC ≥ (3R2 − OH + MH ) − R2 R 2 3 ⇒ MA3 + MB + MC − R.MH ≥ 3R3 − R.OH = const 2 Easily seen equal when M ≡ O Thus we have MA3 + MB + MC − R.MH has minimum value iff M ≡ O Problem Given are ABC with sides a, b, c and A B C with sides a , b , c and area S With any M on plane prove that b2 c2 a2 MA + MB + MC ≥ 4S a b c Trn Quang Hng - High school for gifted students at science Proof We well know the inequality: Given triangle ABC and ∀x, y, z > then (x + y + z)2 ≥ yz sin2 A + zx sin2 B + xy sin2 C we can replace yz → x, zx → y, xy → z we will get the inequality: ( yz + x Now we let zx + y xy ) ≥ x sin2 A + y sin2 B + z sin2 C(∗) z MBMC MBMC = bc.a bc.4R sin2 A MCMA MCMA y= = ca.b ca.4R sin2 B MAMB MAMB z= = ab.c ab.4R sin2 C x= Thus we will have: MCMA MAMB ca.b ab.c = a MA MBMC a.b c x= bc.a yz = x similar we have zx b = MB, y b.c a xy c = MC z c.a b and using inequality (∗) for triangle A B C and x, y, z as above we will get a ( MA)2 ≥ a,b,c a.b c MBMC MBMC sin2 A = ( ) 2 4R bc bc.4R sin A a,b,c a,b,c If we use well know inequality MB.MC MC.MA MA.MB + + ≥1 bc ca ab then we get the consequence inequality: a 1 ( MA)2 ≥ a,b,c a.b c 4R a b c MA + MB + MC ≥ a.b c b.c a c.a b R b2 c2 abc a2 MA + MB + MC ≥ = 4S ⇔ a b c R ⇔ Easily seen we have equal when ABC ∼ ABC and M ≡ H(Orthocenter of triangle ABC) Trn Quang Hng - High school for gifted students at science Remark This inequality have some nice applycations • If ABC ≡ ABC we get the well know inequality aMA + bMB + cMC ≥ 4S • If A B C ≡ Ja Jb Jc with Ja , Jb , Jc are three excenter of triangle ABC with noitice a = B C 2SR A we will get the nice inequality 4R cos , b = 4R cos , c = 4R cos and S = 2 r cot • If ABC ≡ A B C MA + cot MB + cot MC ≥ a + b + c 2 BCA we will get the non symmetry inequality c2 a2 b2 MA + MB + MA ≥ 4S a b c There are some nice other inequality is a consequence of this inequatlity Problem Let M be an arbitrary point inside equaliteral triangle ABC.Find value of 1 + + MA MB MC Proof We can assume ABC is an equaliteral triangle with side 1,let barycentric coordinates of M is (x, y, z), x + y + z = because M inside triangle ⇒ x, y, z > By distance formula we have yz + zx + xy y+z a2 − a = by x + y + z = and a = ⇒ MA = y + yz + z , MA2 = x + y +√ z (x + y + z)2 similarly MB = z + zx + x2 , MC = x2 + xy + y therefore we need find value of y + yz + z +√ + z + zx + x2 x2 + xy + y when x, y, z > 0, x + y + z = 1, we will solve it with Lagrange multipliers WLOG ≤ x ≤ y ≤ z < Case 1: x = we have to prove √ 1 f (y, z) = + + ≥4+ y + yz + z y z indeed √ √ √ y+z y+z f (y, z) ≥ f ( yz, yz) ≥ f ( , )=4+ 2 Case 2: < x ≤ y ≤ z < F (x, y, z, λ) = x2 + xy + y + λ(x + y + z − 1) Trn Quang Hng - High school for gifted students at science ∂F ∂F ∂F =0, =0, = , then ∂x ∂y ∂z − (x2 + xy + − (x2 2y + x + + xy + y )3 − (z 2x + y y )3 2z + x + zx + x2 )3 2x + z + +λ=0 (z + zx + x2 )3 (y 2y + z +λ=0 + yz + z )3 2z + y + (y +λ=0 + yz + z )3 Adding λ= x+y (x2 + xy + y 2)3 + y+z (y + yz + z )3 + z+x (z + zx + x2 )3 Inserting λ in first we get x (x2 Similarly = + xy + y 2)3 = (x2 + xy + y )3 y z (x2 + xy + y )3 = (y + yz + z )3 (z + zx + x2 )3 (x2 + xy + z )3 Hence y 2(z + zx + x2 )3 = z (x2 + xy + y )3 We put y = ax, z = bx, where ≤ a ≤ b a2 (b2 + b + 1)3 = b2 (a2 + a + 1)3 we get a = b Hence y = z as necessary for critical points in the interior of the region < x, y, z < We have to prove √ 2 + √ ≥ 4+ x2 + xy + y y where x + 2y = g(y) = where √ + √ ≥4+ 3y − 3y + y 1 ≤y≤ Trn Quang Hng - High school for gifted students at science 1 , since x ≤ y ≤ z By differentiation it is easily checked that the absolute minimum of g(y) on √ is + = g(1/2) 1 1 Thus + + get value ⇔ M( , , 0)and others permutation ⇔ M concur three MA MB MC 2 midpoint of three side Problem Suppose a, b, c are sidelengths of a triangle and ma , mb , mc are its medians Prove the inequality √ ma mb mc 3(a + b2 + c2 ) + + ≥ a2 b2 c2 2abc Proof This inequality equivalent ( we have ( ma bc mb ca mc ab + + ) ≥ (a + b2 + c2 )2 a b c ma bc mb ca mc ab + + ) ≥ 3( a b c (mb ca).(mc ab) ) = 3( bc a2 mb mc ) we will prove a2 mb mc ) ≥ (a2 + b2 + c2 )2 ⇔ 4( a2 mb mc ) ≥ (a2 + b2 + c2 )2 Indeed turn into triangle with three side ma , mb , mc we need prove: 3 4m2a b c ≥ (m2a + m2b + m2c )2 ⇔ (2(b2 + c2 ) − a2 )bc ≥ (a2 + b2 + c2 )2 4 by equivalent tranformation we have ⇔ (a − (b − c)2 )(b − c)2 ≥ which is true because a > |b − c|, b > |c − a|, c > |a − b| with any triangle ABC 3( Problem Let triangle ABC and X, Y, Z are arbitrary points on segment BC, CA, AB Prove that 1 + + ≥ SAY Z SBZX SCXY SXY Z Lemma 6.1 Let a, b, c > be positive real numbers then 1 + + ≥ a(1 + b) b(1 + c) c(1 + a) + abc Proof We have (1 + abc)( 1 + + )+3= a(1 + b) b(1 + c) c(1 + a) 1+a + a(1 + b) √ b(c + 1) 3 + ≥ √ abc ≥ 1+b abc So we are done.In fact the ineq could be better and stronger as 1 √ + + ≥ √ a(1 + b) b(1 + c) c(1 + a) abc(1 + abc) Trn Quang Hng - High school for gifted students at science Proof We let CY AZ BX = x, = y, = z, SABC = S, < x, y, z < we will have BC CA AB SAY Z SBZX SCXY = z(1 − y), = x(1 − z), = y(1 − x) S S S Thus we have SXY Z = S −SAY Z −SBZX −SCXY = S −S(z(1−y)+x(1−z)+y(1−x)) = S(xyz−(x−1)(y−1)(x−1)) Thus we need to prove + + ≥ SAY Z SBZX SCXY SXY Z S S S 3S ⇔ + + ≥ z(1 − y) x(1 − z) y(1 − x) xyz + (1 − x)(1 − y)(1 − z) (1 − x)(1 − y)(1 − z) 1+ xyz 1−x 1−y 1−z 1 Now let = a > 0, = b > 0, = c > we will get + + ≥ x y z (a + 1)b (b + 1)c (c + 1)a now replace a → b, b → c, c → a we will get our above lemma + abc ⇔ yz zx xy + + ≥ (1 − y) (1 − z) (1 − x) Problem Given two triangles ABC and A B C with ares S, S resp prove that √ aa + bb + bb ≥ 3SS Proof ( aa )2 ≥ 3( ≥ 24SS bb cc ) = 12SS ( = 12SS − cos(A + A ) ) = 24SS sin A sin A ≥ cos(A − A ) − cos(A + A ) √ ≥ 48SS ⇒ aa + bb + cc ≥ 3SS A+A sin2 A+A In the last inequality we easily seen = π thus they are three angle of a triangle therefore apply the inequality ≥ for them,we are done sin2 A ...Trn Quang Hng - High school for gifted students at science Problem Given are value of sum ABC orthorcenter H, circumradius R, with any M on plane find minimum MA3 + MB + MC − R · MH 2 Proof By. .. of M is (x, y, z), x + y + z = because M inside triangle ⇒ x, y, z > By distance formula we have yz + zx + xy y+z a2 − a = by x + y + z = and a = ⇒ MA = y + yz + z , MA2 = x + y +√ z (x + y +... where x + 2y = g(y) = where √ + √ ≥4+ 3y − 3y + y 1 ≤y≤ Trn Quang Hng - High school for gifted students at science 1 , since x ≤ y ≤ z By differentiation it is easily checked that the absolute