AIS 2025 20th International Symposium on Applied Informatics and Related Areas Proving Olympiad Inequalities by Using Some Unusual Ideas Borbé ly József Óbudai Egyetem, Székesfehérvár, Hungary borbely.jozsef@uni-obuda.hu Abstract— There are a lot of challenging problems concerning inequalities in mathematical olympiads which have an elegant The roots of the polynomial p(z) = z2 + n·z - c are or unorthodox elementary solution. In this paper we will present some of these. −n+√n2+4c n2 n = √c + - 2 4 2 2 2 I. INTRODUCTION -n−√n +4c n nand = -√c + - . 2 4 2 In this paper we will prove six olympiad level inequalities 2 using some unusual techniques. Every proof will be Thus p(z) = z + n·z - c = completely elementary. 2 2 = [z − (√c + n − n)]·[z − (−√c + n − n)] 4 2 4 2 II. THE PROBLEMS AND THEIR SOLUTIONS √ n 2 n Here the factor [z − (− c + − )] must be 4 2 First we will prove an inequality, which was proposed in nonnegative, because the value of z = ∑𝑛 𝑥𝑖 is 𝑖=1 2010 by the author in the Hungarian competition NMMV. nonnegative. n2 n Inequality 1: Let n be a positive integer number, and let c Thus [z − (√c + − )] must be nonnegative as well 2 be a positive constant. Iƒ the nonnegative numbers 4 which implies the desired result.⬛ x1, x2, ..... , xn satisƒy the condition Our second inequality is also a problem which was invented ∑ n 2i=1 (xi + ixi)=c, then the inequality by the author and it was posed in the Hungarian competition OKTV in 2012. 2 √c + n - n ≤ ∑n x holds. 4 2 i=1 i Inequality 2: Let 1 ≤ k ≤ n be integer numbers, and let x1, x2,..... , xn be nonnegative numbers with sum Proof: ∑ n i=1 x i = 1. The most important question is how we can use the condition that the numbers are nonnegative. After some With these conditions the maximal possible value oƒ experimentation we can come up with the (maybe surprising) idea that the nonnegativity of the given numbers x1x2 · … · xk + x2x3 · … · xk+1+...+xn−k+1xn−k+2 · … · xn implies the inequality is 1 . k k ∑ n i=1 (xi2 + ix n 2 n i) ≤ (∑i= 1 xi) + n·∑i= 1 xi. This is comfortable for us because the right hand side of the inequality is a second degree polynomial of (∑ni= 1x i). Proof: Let us denote ∑ni= 1x i by ∑ n i=x1 i = z. Using this notation we Here we will use a similar technique to the one we have seen have to check for which values of z we have in the solution of the first problem, namely we will smuggle c ≤ z2 + n·z. some terms into the expression to be able to tackle the problem more easily. This inequality is equivalent to If we consider the expression 0 ≤ z2 + n·z - c. x1x2 · … · xk + x2x3 · … · x k+1+...+xn−k+1xn−k+2 · … · xn, 130 / 243 DOI: 10.12700/AIS.2025.024 AIS 2025 20th International Symposium on Applied Informatics and Related Areas then we can note that in every term the k indices cover all The following problem was also posed in the Hungarian possible remainders if we divide them by k. This competition OKTV in the year 2008. By using a clever observation can us inspire to introduce the following partial substitution we can give a short and elementary solution. sums: for every 0 ≤ r ≤ k-1, let Sr be the sum of the numbers xi for which the index i has the remainder r if we divide it by Inequality 4: k. Using this notation we have Iƒ the sum oƒ the nonnegative real numbers t1, t2, t3 is ∑ n x k−1 t1+t2+t3 = 2π, then i=1 i = ∑r=0 Sr = 1, and 2cost1 + 6cost + 3cost ≥ -7. x1x2 · 2 3 … · xk + x2x3 · … · xk+1+...+xn−k+1xn−k+2 · … · xn ≤ ≤ ∏k−1r=0 S r. Proof: Using the well known inequality between the arithmetic and geometric means we conclude that the value ∏k−1r=0 Sr will be The trick is to interprete the problem geometrically. Let us maximal if S0 = S1 = ... = Sk−1 = 1, thus the value of define the vectors v , v , v in such a way that the angle k 1 2 3 x1x2 · … · xk + x2x3 · … · xk+1+...+xn−k+1xn−k+2 · … · xn between v1 and v2 is t1, the angle between v2 and v3 is t2, the angle between v3 and v1 is t3, and the length of v1 is 1, cannot be greater than 1 . the length of v2 is 2, the length of v3 is 3. k k We will use the dot products of these vectors. Obviously The value 1 can be achieved for example by choosing v v = 2cost , v v = 6cost , v v = 3cost . kk 1 2 1 2 3 2 3 1 3 x1 = x2 = ... = xk = 1, and xk+1 = xk+2 = ... = xn = 0. ⬛ k Thus we have Inequality 3 will be a problem of the Hungarian competition 0 ≤ (v1 + v2 + v 23) = OKTV from the year 2013. Here one can come up with a surprising and elegant solution. = v12 + v22 + v32 + 2·(v1v2 + v2v3 + v3v1) = Inequality 3: = 1+4+9+2·(2cost1 + 6cost2 + 3cost3). For real numbers This implies that a1 ≤ a2 ≤ ... ≤ an ≤ b1 ≤ b2 ≤ ... ≤ bn we have 2 -7 ≤ 2cost1 + 6cost2 + 3cost3. (∑n a + ∑n b ) ≥ 4n·∑n a b . i=1 i j=1 j k=1 k k The value (-7) can be achieved by choosing for example t1 = t2 = π and t3 = 0. ⬛ Proof: Inequality 5 is a problem from the IMO (International Obviously it is enough to prove the inequality Mathematical Olympiad) team selection test held in Hungary from the year 2005. No solution was published for 2 (∑n a + ∑n b ) - 4n·∑n a b ≥ 0. this problem, the following one is the proof of the author. i=1 i j=1 j k=1 k k Inequality 5: The key observation is that the expression on the left hand side is similar to the discriminant of a quadratic polynomial. Let us namely define the polynomial Iƒ a, b, c, d are nonnegative real numbers such that p(z) = ∑n (z − ak) · (z − bk). 2 2 2 2 k=1 a -ab+b = c -cd+d , then (a+b)∙(c+d) ≥ 2∙(ab+cd). It is easy to check that its discriminant is Proof: 2 (∑n a + ∑n b ) - 4n·∑n a b . Since a, b, c, d are nonnegative, they can be considered as i=1 i j=1 j k=1 k k lengths of some segments. Due to the arrangement of our numbers, the value of p(an) Consider a triangle with one side of length “a”, and another side cannot be greater than 0, and the value of p(bn) cannot be of length “b” with an enclosed angle of 60 degrees. less than 0. Thus the polynomial p(z) has a real root, hence Similarly, consider a triangle with one side of length “c”, its discriminant is nonnegative, and this gives the desired another side of length “d”, with an enclosed angle of 60 result. ⬛ 131 / 243 AIS 2025 20th International Symposium on Applied Informatics and Related Areas degrees. The lengths of the third sides in the two triangles - 3∙ sin v - 3∙ sin w. 4 are the same due to the law of cosines. 4 Therefore, due to the generalized law of sines, the radii of Using that sin2(v) ≤ sin v and sin2(w) ≤ sin w we have the the circumcircles of the two triangles are the same. Let the following upper estimate: remaining two angles in the first triangle be (60 ° - φ ) and (60 °+ φ ), respectively, and in the other one (60 ° - µ) and (60 sin2(v) 2(w) sin °+ µ). + sin v + + sin w - 3 - 3∙ sin v∙ sin w - 2 2 4 4 Since the radii of the circumscribed circles are equal, by - 3∙ sin v - 3∙ sin w ≤ normalizing the original inequality using the law of sines, 4 4 we should prove the inequality ≤ sin v + sin v + sin w + sin w - 3 - 3∙ sin v∙ sin w - 2 2 4 4 [sin(60° − φ) + sin(60° + φ)]∙[sin(60° − µ) + sin60 °+ µ ≥ - 3∙ sin v - 3∙ sin w = 4 4 ≥ 2∙ sin(60 ° − φ) ∙ sin(60 ° + φ) + 3 3 3 3 + 2∙ sin(60 ° − µ) ∙ sin(60 ° + µ) = - - ∙ sin v∙ sin w + ∙ sin v + ∙ sin w = . 4 4 4 4 By using the well known identities = 3∙(sin v − 1)∙(1 − sin w). 4 sin(x − y) + sin(x + y) = 2sin x∙ cos y and Here sin v − 1 ≤ 0 and 1 − sin w ≥ 0, thus we proved the statement of the problem. ⬛ cos(x − y) - cos(x + y) = 2sin x∙ sin y Inequality 6 was published in the journal “Matematika the inequality is equivalent to tanítá sa” in 2011. This is a generalization of a problem posed in the competition USAMO in 1980. Here we present the 2∙sin 60°∙cos φ ∙ 2∙sin 60°∙cos µ ≥ solution of the author of this paper, which differs from the official one. ≥ cos(2φ) - cos 120° + cos(2µ) - cos 120°. Inequality 6: This is equivalent to 3∙ cos φ ∙cos µ ≥ cos(2φ) + cos(2µ) +1, Iƒ r1, r2, ..., rn are real numbers and S=∑n 2i =1 sin ri, then which can be written as sin2r ∑n i ∏n 2 i=1 + 2 i=1 cos r i ≤ 1. S+cos ri 0 ≥ 2∙cos2(φ) + 2∙cos2(µ) - 1 - 3∙ cos φ ∙cos µ. Proof: Both φ and µ lie between 0° and 60°, thus the values of cosφ and cos µ lie both between 1 and 1. Thus for For every index 1 ≤ i ≤ n, let xi be xi = sin2ri. Using this 2 notation we have to prove that appropriate real numbers v and w we have 1 1 n ∑ xi (1 − x ) ≤ 1. i=1 + ∏ n i cos φ = ∙ (1 + sin v), cos µ = ∙ (1 + sin w). 1+S−x i=1 i 2 2 We substitute these ones: The role of the numbers 𝑥𝑖 is symmetrical, thus we can assume that 2∙cos2(φ) + 2∙cos2(µ) - 1 - 3∙ cos φ ∙cos µ = 2 0 ≤ x1 ≤ x2 ≤ ... ≤ xn. 1 ( ) 12 ( ) = 2∙[ ∙ 1 + sin v ] + 2∙[ ∙ 1 + sin w ] -1 - Now for every index 1 ≤ i ≤ n we have 2 2 -3∙ [1 ∙ (1 + sin v)] ∙[1 ∙ (1 + sin w)] = x x 2 2 i ≤ i . 2 1+S−xi 1+x1+x2+⋯+x𝚗−1 (v) 2(w) = 1 + sin + sin v + 1 + sin + sin w -1 - 3 - 2 2 2 2 4 Summing up these inequalities we get - 3∙ sin v∙ sin w - 3∙ sin v - 3∙ sin w = n x∑ i + ∏ n (1 − x ) ≤ 4 4 4 i=1 1+S−x i=1 i i 2(v) 2(w) = sin + sin v + sin + sin w - 3 - 3∙ sin v∙ sin w - ≤ ∑n xi + ∏n (1 − x )= 2 2 4 4 i=1 1+x +x i=1 i 1 2+⋯+x𝚗−1 132 / 243 AIS 2025 20th International Symposium on Applied Informatics and Related Areas = 1 - 1−x𝚗 +(1 − xn) · ∏n−1(1 − xi) = 1+x1+x2+⋯+x i=1 𝚗−1 = 1 - (1 − x )·( 1 n − ∏n−1(1 − xi)) . 1+x1+x2+⋯+x i=1 𝚗−1 Since 1 − xn ≥ 0, it would be enough to prove that 1 − ∏n−1i=1( 1 − xi) ≥ 0, which is equivalent to 1+x1+x2+⋯+x𝚗−1 (1 + x1 + x2 + ⋯ + xn−1)·∏n−1i=1 (1 − xi) ≤ 1. Due to the nonnegativity of the numbers 𝑥𝑖 we have (1 + x1 + x2 + ⋯ + x ) ≤ ∏n−1n−1 i=1 (1 + xi). Using this observation we get (1 + x1 + x n−12 + ⋯ + xn−1)·∏i=1 (1 − xi) ≤ ≤ ∏n−1(1 + x )·∏n−1i (1 − xi) = i=1 i=1 = ∏n−1(1 − xi2) ≤ ∏n−1 1 = 1, i=1 i=1 which is the desired result. ⬛ REFERENCES [1] Duš an Djukić , Vladimir Janković , Ivan Matić , Nikola Petrović , The IMO Compendium: A Collection of Problems Suggested for The International Mathematical Olympiads: 1959-2009 Second Edition, Problem Books in Mathematics, 2nd ed., 2011 [2] A matematika tanítá sa, Mozaik kiadó, 2011/1. szá m, 419. feladat Online sources: [3] http://benoke98.f.fazekas.hu/olimpia/feladatok/valogato2005.pdf [4] https://nmmv.berzsenyi.hu/feladatok/2010 [5] https://www.oktatas.hu/kozneveles/tanulmanyi_versenyek_/oktv_kerete ben/versenyfeladatok_javitasi_utmutatok 133 / 243