One group of inequalities with altitudes and medians in triangle

TRANSACTIONS OF THE AMERICAN MA THEMA TICAL SOCIETY V olume 00, N umber 0, Pages 000– 000 S 0002-9947(XX)0000 -0 ONE GR OUP OF INEQUALITIES WITH AL TITUDES AND MEDIANS IN TRIANGLE ZHIVKO ZHELEV Abstract. In the article w e pro ve some inequalities that con tain r elations betw een altitudes and medians in t riangle. At least one of these i nequalities has not been considered in the literature b efore and the m ain theorem has also not been prov ed elsewhere in that form. Some immediate corollaries ha v e been presen ted as well. Geometry of the triangle is a r ealm of elementary geo metry where interested new results pop up all the time. There ar e plent y of theor ems concerning geometry of the triangle, including hundreds of geo metric ineq ualities (see for example [2]). The following result c onsists of tw o inequalities which lo ok quite pleasant but the second o ne turned out to b e e x tremely difficult to tackle. In fac t we prove the following Theorem 1. L et △ AB C b e an arbitr ary triangle with sides a , b and c . L et also h a , h b , h c and m a , m b , m c b e the altitudes and me dians to t hese sides resp e ctively. Then t he fol lowing two ine qualities hold: ah a + bh b + ch c ≤ √ bch a + √ ach b + √ abh c (1) am a + bm b + cm c ≤ √ bcm a + √ acm b + √ abm c . (2) As far a s the author knows, the seco nd inequa lit y has no t b een prov ed yet. But different ra ndom chec ks of v alues of the triangle sides hav e not brought any counterexamples. Our pro of is based on one and tw o v ariable functions theory which we consider an impediment since the pro blem for m ulated above is in the ar ea of the elementary mathematics. First we formulate so me le mma s which are mo r e or less obvious. Lemma 1 ( Arithm etic-Geometric-Mean I nequalit y). L et a 1 , a 2 , . . . a n ∈ R and a i ≥ 0 , i = 1 , . . . , n . Then the fol lowing ine quality is true: (3) a 1 + a 2 + · · · + a n n ≥ n √ a 1 a 1 · · · a n . Pro of. See [4, p. 18-1 9] and [3] for a deta iled pro of.  Lemma 2. If a + b + c > 0 , then a 3 + b 3 + c 3 ≥ 3 ab c . Key wor ds and phr ases. medians, altitudes, inequality . 2000 Mathematics Subje ct Classific ation. 51M04, 51M05, 51M16. c  1997 American Mathematical So ciety 1 2 ZHIVKO ZHELEV Pro of. F r om Lemma 1 ( n = 3 ), we get that a 1 + a 2 + a 3 ≥ 3 3 √ a 1 a 2 a 3 . Replacing a 1 with a 3 , a 2 with b 3 and a 3 with c 3 , we g et that a 3 + b 3 + c 3 ≥ 3 3 √ a 3 b 3 c 3 = 3 ab c ., q. e. d.  Pro of of the main theorem. In order to prove (1) we use that ah a = bh b = ch c = 2 S △ AB C . Then ah a + bh b + ch c − √ bch a − √ ach b − √ abh c = 6 S − 2 S √ bc a − 2 S √ ac b − 2 S √ ab c = 2 S 3 − √ bc a − √ ac b − √ ab c ! . Therefore, ah a + bh b + ch c ≤ √ bch a + √ ach b + √ abh c ⇐ ⇒ 3 − √ bc a − √ ac b − √ ab c ! ≤ 0 ⇐ ⇒ 3 abc ≤ ( bc ) 3 2 + ( ac ) 3 2 + ( ab ) 3 2 , a > 0 , b > 0 , c > 0 . By the substitution x = √ bc , y = √ ac , z = √ ab , we ge t that 3 abc ≤ ( bc ) 3 2 + ( ac ) 3 2 + ( ab ) 3 2 ⇐ ⇒ 3 xy z ≤ x 3 + y 3 + z 3 , x > 0 , y > 0 , z > 0 . Last inequality is exac tly L e mma 2 and the pro of of (1 ) is completed. Now, without loss o f genera lit y , w e can ass ume for △ AB C , tha t a ≥ b ≥ c . Three main cases ar e p ossible: † ) △ AB C is equilateral, i. e. a = b = c ; †† ) △ AB C is isosceles, i. e. a = b > c ; † † † ) △ AB C is an a rbitrary triangle , i. e. a > b > c . First, let’s rewr ite (2 ) in the form (4) ( a − √ bc ) m a + ( b − √ ac ) m b + ( c − √ ab ) m c ≤ 0 . First case. If △ AB C is equila teral, then a − √ bc = b − √ ac = c − √ ab = 0 and (4) is fulfilled as an equality . W e will see b elow that this ca s e is in fact the extremal case for our pro blem. Second case. Now let △ AB C b e iso s celes and we ma y a ssume a = b > c > 0. On the other hand we hav e that (5) m a = 1 2 √ 2 b 2 + 2 c 2 − a 2 = 1 2 √ a 2 + 2 c 2 = m b , m c = 1 2 √ 2 a 2 + 2 b 2 − c 2 = 1 2 √ 4 a 2 − c 2 . Then using (5), (4) is transformed in to ONE GR OUP OF INEQUALITIES 3 (6) ( a − √ ac ) p a 2 + 2 c 2 ≤ a − c 2 p 4 a 2 − c 2 , a > c. W e will prov e (6). After so me tedio us computations, we get co nsequently: ( a − √ ac ) √ a 2 + 2 c 2 ≤ a − c 2 √ 4 a 2 − c 2 ⇐ ⇒ √ a ( √ a − √ c ) √ a 2 + 2 c 2 ≤ ( √ a − √ c )( √ a + √ c ) 2 √ 4 a 2 − c 2 ⇐ ⇒ 2 √ a 3 + 2 ac 2 ≤ ( √ a + √ c ) √ 4 a 2 − c 2 ⇐ ⇒ 4( a 3 + 2 ac 2 ) 4 a 2 − c 2 ≤ a + c + 2 √ ac ⇐ ⇒ 4 a 3 + 8 ac 2 − 4 a 3 + ac 2 − 4 a 2 c + c 3 4 a 2 − c 2 ≤ 2 √ ac ⇐ ⇒  c 3 + 9 ac 2 − 4 a 2 c 4 a 2 − c 2  2 ≤ 4 ac ⇐ ⇒ c 6 + 81 a 2 c 4 + 16 a 4 c 2 + 18 ac 5 − 8 a 2 c 4 − 72 a 3 c 3 ≤ 64 a 5 c + 4 ac 5 − 32 a 3 c 3 ⇐ ⇒ c 6 + 14 ac 5 + 73 a 2 c 4 − 40 a 3 c 3 + 16 a 4 c 2 − 64 a 5 c ≤ 0 ⇐ ⇒ c ( c 5 + 14 ac 4 + 73 a 2 c 3 − 40 a 3 c 2 + 16 a 4 c − 64 a 5 ) ≤ 0 ⇐ ⇒ c 5 + 14 ac 4 + 73 a 2 c 3 − 40 a 3 c 2 + 16 a 4 c − 64 a 5 ≤ 0 ⇐ ⇒  c a  5 + 14  c a  4 + 73  c a  3 − 40  c a  2 + 16  c a  − 64 ≤ 0 . In the last expr ession, le t c a = t ∈ (0 , 1 ), a nd it follows that t 5 + 14 t 4 + 73 t 3 − 40 t 2 + 16 t − 64 ≤ 0 ⇐ ⇒ ( t − 1) | {z } < 0 ( t 4 + 15 t 3 + 88 t 2 + 48 t + 64) | {z } > 0 if t> 0 ≤ 0 . But the last one is obviously true and that prov e s (6) in this ca se.  Third case. Let now △ AB C b e an a rbitrary triang le and let a > b > c > 0. W e rewrite (4) in the for m (7) 1 2 ( a − √ bc ) p 2 b 2 + 2 c 2 − a 2 + 1 2 ( b − √ ac ) p 2 a 2 + 2 c 2 − b 2 + 1 2 ( c − √ ab ) p 2 a 2 + 2 b 2 − c 2 ≤ 0 or 4 ZHIVKO ZHELEV (8) 1 2 a 2   1 − r b a r c a ! s 2  b a  2 + 2  c a  2 − 1 +  b a − r c a  s 2 + 2  c a  2 −  b a  2   | {z } F 0 @ b a , c a 1 A := F ( x ,y ) + 1 2 a 2   c a − r b a ! s 2 + 2  b a  2 −  c a  2   | {z } F 0 @ b a , c a 1 A := F ( x ,y ) ≤ 0 , and therefore 1 2 a 2 F ( x, y ) ≤ 0 ⇐ ⇒ F ( x, y ) ≤ 0. Two v ariable function F ( x, y ) defined ab ov e, w e na me devil-fish function and the surface this function plots in R 3 – devil-fish surfac e . In our new notation inequa lit y (4) transfor ms into the ex tr emal problem (9) max ( x,y ) ∈ M ⊂ R 2 F ( x, y ) , where M = { ( x, y ) ∈ R 2 | 0 ≤ y ≤ x ≤ 1 , x + y ≥ 1 } , where M is geometrically a r ight-angle triangle and the last inequalit y follo ws from the fact that a + b > c in an arbitrary triangle. It is a straig htforward chec k that the devil-fish function is well-defined on M . Therefore, in order to prov e (4), we have to prov e that max ( x,y ) ∈ M F ( x, y ) = 0 . Note tha t M is a compact set and since M is a co ntin uous function on M , it follows from some cla ssical results in the real mathematical a na lysis, that F ( x, y ) reaches its maximum there. In o rder to find that max ima l p oint we consider the int erior and the b oundary of M separ ately , M = int M ∪ ∂ M . Using s ome c a lculating pro grams such as Maple (computations can be done manually but that can turn into an extremely tedious hardwork), we find that ∂ F ( x, y ) ∂ x = − y p 2 x 2 + 2 y 2 − 1 2 √ xy + 2(1 − √ xy ) x p 2 x 2 + 2 y 2 − 1 + p 2 + 2 y 2 − x 2 − ( x − √ y ) x 2 + 2 y 2 − x 2 − 2 + 2 x 2 − y 2 2 √ x + 2( y − √ x ) x p 2 + 2 x 2 − y 2 , and since the devil- fish function is a symmetric function, i. e. F ( x, y ) = F ( y , x ), it follows immediately that ONE GR OUP OF INEQUALITIES 5 ∂ F ( x, y ) ∂ y = − x p 2 y 2 + 2 x 2 − 1 2 √ xy + 2(1 − √ xy ) y p 2 y 2 + 2 x 2 − 1 + p 2 + 2 x 2 − y 2 − ( y − √ x ) y 2 + 2 x 2 − y 2 − 2 + 2 y 2 − x 2 2 √ y + 2( x − √ y ) y p 2 + 2 y 2 − x 2 . Then w e solve the sys tem          ∂ F ∂ x = 0 ∂ F ∂ y = 0 , which gives us tw o p oints: M 1 (0 , 9238 12749 1 . . . , 0 , 16 60179 102 . . . ) and M 2 (1 , 1). Note that M 1 , M 2 ∈ M and M 1 ∈ int M , M 2 ∈ ∂ M . Again using some softw are, we get for the devil- fish function’s hessian: H ( x, y ) := ∂ 2 F ∂ x 2 · ∂ 2 F ∂ y 2 −  ∂ 2 F ∂ x∂ y  2 | M 2 (1 , 1) = − 3 √ 3 2 ! − 3 √ 3 2 ! − 3 √ 3 4 ! 2 =  9 4  2 > 0 and s ince ∂ 2 F ∂ x 2 | M 2 (1 , 1) = − 3 √ 3 2 < 0, it follows that p oint M 2 (1 , 1) is a po ssible po int of max imum over M . F o r completeness we nee d to chec k wha t is go ing on the b oundar y b ecaus e M 2 ∈ ∂ M . But that is straig h tforward and this chec king consists of the cases { x = 0 } , { x = y } , { x = 1 } , and { x + y = 1 } . This leads to the fact that the p oint in question is in fact an abso lute maximum in M . This yields sup ( x,y ) ∈ M F ( x, y ) = max ( x,y ) ∈ M F ( x, y ) = F (1 , 1) = 0 . Additionally , M 1 is a p oint of a lo c al minim um and more over: F ( x, y ) | M 1 = F (0 , 92 38127 491 . . . , 0 , 16601 79102 . . . ) = − 0 , 428 0 65796 8 . . . 6 ZHIVKO ZHELEV The devil-fish surfa c e can be seen b elow: 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 -0.6 -0.4 -0.2 0 0 0.2 0.4 0.6 0.8 Figure 1. Devil-fis h sur face Using soft w are such a s Maple and Mathematic a , one can get different v iews to that surface: ONE GR OUP OF INEQUALITIES 7 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 -0.6 -0.4 -0.2 0 (a) View from ab ov e 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 -0.6 -0.4 -0.2 0 0 0.2 0.4 0.6 0.8 1 (b) Left corner view 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 -0.6 -0.4 -0.2 0 0 0.2 0.4 0.6 0.8 1 0 (c) Ri gh t corner view 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 (d) F ron t view 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 -0.6 -0.4 -0.2 0 0 1 0. 2 0 .4 0.6 0.8 1 (e) F ront -do wn view 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 -0.6 -0.4 -0.2 0 0 0.2 0.4 0.6 0.8 1 0. 2 0 .4 0.6 0.8 1 (f ) F ront -up view Figure 2. Different views of the devil- fish surface Therefore all the cases a re do ne and the theore m is prov ed.  The following immediate co rollar y is true: Corollary 1 . L et a, b and c b e the sides of a triangle △ AB C and let also m a , m b m c b e the me dians to these sides r esp e ctively. Then a) (2 p − 3 a ) m a + (2 p − 3 b ) m b + (2 p − 3 c ) m c ≥ 0 , whe r e p := a + b + c 2 . b) m a m c ≤ √ ab + √ ac + √ bc a + b + c ≤ 1 , a ≥ b ≥ c . References [1] G. Boich ev. Geometric inequalities containing medians and some other elements of the tri- angle. Educ ation in mathematics , 1:41–49, 1981. (in Bulgarian). [2] O. Bottema, R. ˆ Z. Djordevi´ c, R. R. Jani ´ c, D. S. Mitri no vi ´ c, and P . M. V asi´ c. Ge ometric ine q ualities . Groningen, 1969. [3] R. B. Nelson. Pro of Without W ords: The Arithmetic-Logarithmic-Geometric Mean Inequal- ity . Math. Mag. , 8:305, 1995. [4] T. Stoilo v and K. Chilingirov a. Ine quality pr oblems . Naro dna prosve ta, Sofia, 1989. (in Bul- garian). Zhivko Zhelev Dep ar tment of Ma them a tics and Informa tics , Trakia University, Rektora t, AF, 367A, 6003 St. Zagora, Bulgaria. Email: zhelev@u ni-sz.bg