HUNGOnsometriangleinequalities

<span class='text_page_counter'>(1)</span>On some triangle inequalities Nguyen Viet Hung High School For Gifted Students, Hanoi University of Science, Vietnam Email address: Abstract. In this article we will use two known triangle inequalities to give some other results.. 1. Introduction. The following two inequalities have been very popular. x2 + y 2 + z 2 ≥ yz. cos A + zx. cos B + xy. cos C 2. (1). x2 + y 2 + z 2 A B C ≥ yz. sin + zx. sin + xy. sin (2) 2 2 2 2 where A, B, C are three angles of a triangle and x, y, z are any real numbers. We can rewrite these two inequalities as b2 + c2 − a2 c2 + a2 − b2 a2 + b2 − c2 + zx. + xy. bc ca ab √ √ √ x2 + y 2 + z 2 (s − b)(s − c) (s − c)(s − a) (s − a)(s − b) ≥ yz. + zx. + xy. 2 bc ca ab There are two popular proofs for (1) as follows x2 + y 2 + z 2 ≥ yz.. (3). (4). Using properties of vector. Let I be the incenter of triangle ABC, and let X, Y, Z be respectively feets of perpendicular lines from I to sides BC, CA, AB. By full expanding the following selfevident inequality −→ −→ − → (xIX + y IY + z IZ)2 ≥ 0 we get the desired inequality. The equality occurs if and only if −→ −→ − → xIX + y IY + z IZ = ⃗0. On the other hand, according to the porcupine theorem, we have −→ −→ − → aIX + bIY + cIZ = ⃗0. Thus the necessary and sufficient conditions such that the equality occurs as y z x = = . a b c. 1.

<span class='text_page_counter'>(2)</span> HSGS Hanoi. Tel: (+84)0979515086. Using algebraic techniques. The inequality (1) is equivalent to 2yz cos A + 2zx cos B − 2xy cos(A + B) ≤ x2 + y 2 + z 2 , 2yz cos A + 2zx cos B − 2xy cos A cos B ≤ x2 + y 2 + z 2 − 2xy sin A sin B, 2yz cos A + 2zx cos B − 2xy cos A cos B + y 2 sin2 A + x2 sin2 B ≤ x2 + y 2 + z 2 + (y sin A − x sin B)2 , 2yz cos A + 2zx cos B − 2xy cos A cos B ≤ x2 cos2 B + y 2 cos2 A + z 2 + (y sin A − x sin B)2 , 2z(y cos A + x cos B) ≤ (y cos A + x cos B)2 + z 2 + (y sin A − x sin B)2 , (y cos A + x cos B − z)2 + (y sin A − x sin B)2 ≥ 0 Which is obviously true. The equility occurs iff { y x sin A = sin B , z = y cos A + x cos B. x y = = k. Then x = k. sin A, y = k. sin B and z = sin A sin B k(sin A cos B + sin B cos A) = k. sin C. So, the conditions above are equivalent to To do more clearly, we set. y z x = = . sin A sin B sin C This is the necessary and sufficient conditions to the equality happens. Applying (1) for a triangle which has three angles. B+C C+A A+B 2 , 2 , 2. we obtain (2).. Remark 1. From the second proof we observe that for any angles α, β, γ (they are not necessary three angles of a triangle) such that α + β + γ = π, the inequality below is also true (for all real numbers x, y, z) x2 + y 2 + z 2 ≥ 2yz cos α + 2zx cos β + 2xy cos γ. (5). 2. Some results. In all problems below, we use known notations of triangle ABC and note that S denotes its area. √ √ √ In (3) we replace (x, y, z) by ( a, b, c) to yield a+b+c≥. b2 + c2 − a2 c2 + a2 − b2 a2 + b2 − c2 √ √ √ + . + ca bc ab. This inequality has equivalent forms as a2 b2 c2 √ + √ + √ +a+b+c≥ ca bc ab. √. a3 + b. √. b3 + c. √. c3 + a. √. b3 + a. √. c3 + b. √. cot A cot B cot C 1 √ + √ ≤ . + √ 2r ca bc ab If we substitue (a2 , b2 , c2 ) for (x, y, z), we get a4 + b4 + c4 ≥ bc(b2 + c2 − a2 ) + ca(c2 + a2 − b2 ) + ab(a2 + b2 − c2 ) or a4 + b4 + c4 + abc(a + b + c) ≥ ab(a2 + b2 ) + bc(b2 + c2 ) + ca(c2 + a2 ). 2. a3 , c.

<span class='text_page_counter'>(3)</span> HSGS Hanoi. Tel: (+84)0979515086. This is Schur’s inequality of fourth degree that is well-known. 1 1 1 We repalace (x, y, z) in (1) by ( √s−a , √s−b , √s−c ) (and for the acute triangle ABC) we have ∑ 1 1 2 1 √ + + ≥ cos A s−a s−b s−c (s − b)(s − c) cyc ≥. 4 4 4 cos A + cos B + cos C. a b c. This can be rewritten by other forms as follows ra + rb + rc cos A cos B cos C ≥ + + . 4S a b c √ √ √ When (x, y, z) is replaced by ( ra , rb , rc ) (the triangle ABC is also acute) we obtain √ √ ra + rb + rc √ ≥ rb rc cos A + rc ra cos B + ra rb cos C 2 ≥ ha cos A + hb cos B + hc cos C Which is the other form of one of the results above. √ √ √ To be continue, we replace again (x, y, z) in (1) by ( rrbarc , rcrrba , rarcrb ) to get rb rc rc ra ra rb + + ≥ 2(ra cos A + rb cos B + rc cos C). ra rb rc In (1),(2) we replace respectively (x, y, z) by ( √1h , √1h , √1h ) and ( √1ra , √1rb , √1rc ) and using a. c. b. 1 1 1 1 1 1 1 + + = + + = ha hb hc ra rb rc r we obtain the following results 1 cos A cos B cos C √ ≤ , +√ +√ 2r hb hc ha hb hc ha sin A sin B sin C 1 √ 2 +√ 2 +√ 2 ≤ . rb rc rc ra ra rb 2r Chosing x =. 1 s−a , y. =. 1 s−b , z. =. 1 s−c. and substitue it into (2) gives. sin A2 sin B2 sin C2 1 + + ≤ (s − b)(s − c) (s − c)(s − a) (s − a)(s − b) 2. (. 1 1 1 + + (s − a)2 (s − b)2 (s − c)2. ) ,. which is equivalent to B C A + (s − b) sin + (s − c) sin ≤ 2 2 2 ( ) (s − a)(s − b)(s − c) 1 1 1 ≤ + + 2 (s − a)2 (s − b)2 (s − c)2. (s − a) sin. On the other hand, it’s easy to show that (s − a)(s − b)(s − c) ≤ We infer that (s − a) sin. A B C abc + (s − b) sin + (s − c) sin ≤ 2 2 2 16 3. (. abc 8. 1 1 1 + + (s − a)2 (s − b)2 (s − c)2. ) ..

<span class='text_page_counter'>(4)</span> From (3) chosing x = (. a b+c. )2. ( +. a b+c , y. b c+a. HSGS Hanoi. =. )2. b c+a , z. ( +. c a+b. c a+b. = )2. ≥. Tel: (+84)0979515086. yields. c2 + a2 − b2 a2 + b2 − c2 b2 + c2 − a2 + + . (b + a)(c + a) (c + b)(a + b) (a + c)(b + c). This inequality has equivalent forms as follows cot A cot B cot C 1 + + ≤ (a + b)(a + c) (b + c)(b + a) (c + a)(c + b) 4S √ Chosing x = √. a b+c , y. √ =. (s − b)(s − c) + (a + b)(a + c). √. b c+a , z. √ =. c a+b. ((. a b+c. )2. ( +. b c+a. )2. ( +. c a+b. )2 ) .. and replace it into (4) we get. (s − c)(s − a) + (b + c)(b + a). √. (s − a)(s − b) 1 ≤ (c + a)(c + b) 2. (. a b c + + b+c c+a a+b. ) .. √ √ √ In (4) we replace (x, y, z) by ( s − a, s − b, s − c) then (s − b)(s − c) (s − c)(s − a) (s − a)(s − b) s √ √ √ + + ≤ . 2 ca bc ab This inequality has equivalent forms as follows √ a+b+c A √ B √ C . bc sin2 + ca sin2 + ab sin2 ≤ 2 2 2 4 1. 1 1 ,y = √ ,z = √ and then (s − b)(s − c) (s − c)(s − a) (s − a)(s − b) respectively substitue it into (1), (4) and note that Chosing again x = √. 1 1 1 1 + + = 2 (s − b)(s − c) (s − c)(s − a) (s − a)(s − b) r we have the following results cos A cos B cos C 1 √ √ √ + + ≤ 2, 2r (s − a) (s − b)(s − c) (s − b) (s − c)(s − a) (s − c) (s − a)(s − b) 1 1 1 1 √ + √ + √ ≤ 2. 2r (s − a) bc (s − b) ca (s − c) ab √ √ √ sin B2 sin C2 sin C2 sin A2 sin A2 sin B2 Now we chose x = , y = , z = and substitue it into (2) sin A2 sin B2 sin C2 to get sin B2 sin C2 sin C2 sin A2 sin A2 sin B2 A B C + + ≥ 2 sin2 + 2 sin2 + 2 sin2 . A B C 2 2 2 sin 2 sin 2 sin 2 (√ ) √ √ s−a s−b s−c When replace (x, y, z) by , , into (4) we have a b c 1 2. (. s−a s−b s−c + + a b c. ) ≥. (s − b)(s − c) (s − c)(s − a) (s − a)(s − b) + + bc ca ab. This is equivalent to A B C s−a s−b s−c + + ≥ 2 sin2 + 2 sin2 + 2 sin2 a b c 2 2 2 4.

<span class='text_page_counter'>(5)</span> HSGS Hanoi. (. or (a + b + c). 1 1 1 + + a b c. Tel: (+84)0979515086. ) + 2(cos A + cos B + cos C) ≥ 12,. ) 2r 1 1 1 + + + ≥ 10. a b c R √ √ √ The next, we also replace (x, y, z) by ( bc, ca, ab) into (4) then obtain (. or. (a + b + c). √ √ √ ab + bc + ca . a (s − b)(s − c) + b (s − c)(s − a) + c (s − a)(s − b) ≤ 2 a b c ,y = ,z = then s−a s−b s−c ( )2 ( )2 ( )2 a b c 2 2 2 + + ≥ + + ≥ 12. A B s−a s−b s−c sin 2 sin 2 sin C2. In (4) we chose x =. Now we replace again (x, y, z) by (sin A′ , sin B ′ , sin C ′ ), where A, B, C are three angles of any triangle, into (1), we find that ∑ cyc. 1 sin B ′ sin C ′ cos A ≤ (sin2 A′ + sin2 B ′ + sin2 C ′ ) 2. Dividing both sides of this inequality by sin A′ sin B ′ sin C ′ , we get ( ) cos A sin A′ cos B cos C 1 sin B ′ sin C ′ + + ≤ + + sin A′ sin B ′ sin C ′ 2 sin B ′ sin C ′ sin C ′ sin A′ sin A′ sin B ′ Note that sin A′ sin(B ′ + C ′ ) sin B ′ cos C ′ + cos B ′ sin C ′ ) = = sin B ′ sin C ′ sin B ′ sin C ′ sin B ′ sin C ′ ′ = cot B + cot C ′ Similarly. sin B ′ = cot C ′ + cot A′ sin C ′ sin A′ sin C ′ = cot A′ + cot B ′ sin A′ sin B ′ Thus, we have a result: For any two triangles ABC and A′ B ′ C ′ , the following inequality holds cot A + cot B + cot C ≥. cos A′ cos B ′ cos C ′ + + . sin A sin B sin C. If we replace again (x, y, z) by (a′ , b′ , c′ ) into (1) and using formula a′2 +b′2 +c′2 = 4S ′ (cot A′ + cot B ′ + cot C ′ ) then to obtain 2S ′ (cot A′ + cot B ′ + cot C ′ ) ≥ b′ c′ cos A + c′ a′ cos B + a′ b′ cos C. This is equivalent to cot A′ + cot B ′ + cot C ′ ≥ Thus, we get again the above result. 5. cos A cos B cos C + + . ′ ′ sin A sin B sin C ′.

<span class='text_page_counter'>(6)</span> HSGS Hanoi. Tel: (+84)0979515086. We consider any point P in triangle ABC and let P X, P Y, P Z be internal bisectors of ∠BP C, ∠CP A, ∠AP B, respectively We set ∠BP C = 2α, ∠CP A = 2β, ∠AP B = 2γ. Using the known formulas about the length of bisectors in a triangle, we have PX = or. ( 2 cos α = P X. 2P B.P C 2P C.P A 2P A.P B cos α, P Y = cos β, P Z = cos γ PB + PC PC + PA PA + PB. 1 1 + PB PC. ). ( , 2 cos β = P Y. 1 1 + PC PA. ). ( , 2 cos γ = P Z. 1 1 + PA PB. ) .. Applying (5) for α, β, γ which are determined above, we have ( ) ( ) ( ) 1 1 1 1 1 1 PX + yz + P Y + zx + P Z + xy ≤ x2 + y 2 + z 2 . PB PC PC PA PA PB √ √ √ We continue take x = P A, y = P B, z = P C then to get (√ ) (√ ) (√ ) √ √ √ PC PB PA PC PB PA +√ + PY √ +√ + PZ √ +√ ≤ P A + P B + P C. PX √ PB PC PC PA PA PB which implies that P A + P B + P C ≥ 2(P X + P Y + P Z) ≥ 2(P Pa + P Pb + P Pc ) where Pa , Pb , Pc are feet of perpendicular lines from P to the sides BC, CA, AB, respectively. We have just received a result which is stronger than Erdos-Mordell inequality. We have known that if a, b, c are side-lengths of a triangle then there exist positive real numbers u, v, w such that a = v + w, b = w + u, c = u + v which is called Ravi’s substitutions. Using this substitutions, (3) can be written as x2 + y 2 + z 2 ≥. ∑. yz ·. (w + u)2 + (u + v)2 − (v + w)2 (w + u)(u + v). yz ·. 2(u2 + uv + uw − vw) (w + u)(u + v). cyc. =. ∑ cyc. ∑. 2(u + v)(u + w) − 4vw (u + v)(u + w) cyc ) ∑ ( 4vw = yz 2 − (u + v)(u + w) cyc. =. yz ·. It follows that yz. vw wu uv 2(xy + yz + zx) − (x2 + y 2 + z 2 ) + zx + xy ≥ (u + v)(u + w) (v + w)(v + u) (w + u)(w + v) 4 2 (x + y + z) − 2(x2 + y 2 + z 2 ) = 4. Note that the equality occurs iff x y z = = v+w w+u u+v 6.

<span class='text_page_counter'>(7)</span> HSGS Hanoi. Tel: (+84)0979515086. We choose x = y = z to get the known result vw wu uv 3 + + ≥ (u + v)(u + w) (v + w)(v + u) (w + u)(w + v) 4 When choosing x = 1, y = 12 , z =. 1 3. then we obtain. vw 2wu 3uv 23 + + > (u + v)(u + w) (v + w)(v + u) (w + u)(w + v) 24 (because the equality does not occur). Applying (3) for a triangle which has three side-lengths ma , mb , mc we get yz. 5b2 − c2 − a2 5c2 − a2 − b2 5a2 − b2 − c2 + zx + xy ≤ 4(x2 + y 2 + z 2 ) mb mc mc ma ma mb. A simple consequent of this result as 5a2 − b2 − c2 5b2 − c2 − a2 5c2 − a2 − b2 + + ≤ 12. mb mc mc ma ma mb In (3) we replace (x, y, z) by (xa, yb, zc) to give x2 a2 + y 2 b2 + z 2 c2 ≥ yz(b2 + c2 − a2 ) + zx(c2 + a2 − b2 ) + xy(a2 + b2 − c2 ) which is equivalent to a2 (x2 + 2yz) + b2 (y 2 + 2zx) + c2 (z 2 + 2xy) ≥ (a2 + b2 + c2 )(xy + yz + zx). √ √ √ The equality holds for x = y = z. In (4) we replace (x, y, z) by (x a, y b, z c) to yield yz. √. (s − b)(s − c) + zx. √. (s − c)(s − a) + xy. √ x2 a + y 2 b + z 2 c (s − a)(s − b) ≤ . 2. From here, choosing again (x, y, z) = ( a1 , 1b , 1c ) we have √. (s − b)(s − c) + bc. √. (s − c)(s − a) + ca. √. (s − a)(s − b) 1 ≤ ab 2. (. 1 1 1 + + a b c. ) .. By similar ways, you can also establish new results for yourself. To finish this article, we will give some problems that can be solved by using elementary inequality above.. 3. Proposed problems. Excersise 1 (S.Klamkin). Let ABC be a triangle and let x, y, z be any real numbers. Prove that x2 + y 2 + z 2 ≥ 2(−1)n+1 (yz cos nA + zx cos nB + xy cos nC) where n is a natural number. The equality occurs iff x y z = = . sin nA sin nB sin nC Hint: The desired inequality is equivalent to (x + (−1)n (y cos nC + z cos nB))2 + (y sin nC − z sin nB)2 ≥ 0. 7.

<span class='text_page_counter'>(8)</span> HSGS Hanoi. Tel: (+84)0979515086. Excersise 2. Let ABC be a triangle and let n be a natural number. Prove that (a) cos(2nA) + cos(2nB) + cos(2nC) ≥ −3/2. (b) cos(2n + 1)A + cos(2n + 1)B + cos(2n + 1)C ≤ 3/2. Excersise 3. Prove that for all any triangle ABC, then √ √ 3 cos A + 2 cos B + 2 3 cos C ≤ 4. Excersise 4. Let ABC be a triangle and let x, y, z be any real numbers. Prove that (. ax + by + cz 4S. )2 ≥. xy yz zx + + . ab bc ca. Excersise 5. Prove that for all any triangle ABC, then (. a2 + b2 + c2 4S. )2 ≥. c2 a2 b2 + + . b2 c2 a2. Excersise 6 (IMO Shortlist, 1995). Given positive real numbers a, b, c. Find all triples (x, y, z) of real numbers satisfying the following system of equation { x + y + z = a + b + c, 4xyz − (a2 x + b2 y + c2 z) = abc. Excersise 7 (China TST, 2007). Let x, y, z be positive real numbers such that x+y+z+ Prove that. √. yz + x. √. zx + y. √ xyz = 4.. √. xy ≥ x + y + z. z. Tài liệu [1] Math and Young Journal, Education Publishing House of Vietnam. [2] Titu Andreescu, Zuming Feng, 103 trigonometry problems, Birkhauser, 2004. [3] Dragoslav S. Mitrinovic, J. Pecaric,V. Volenec„ Recent Advances in Geometric Inequalities.. 8.

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