Ravi Substitution
It is natural to require that the inequality in Theorem 2 holds for arbitrary positive reals a, b, c. It is possible to prove the inequality without the additional condition that a,b,c are the lengths of a triangle. Since the inequality is symmetric in the variables, without loss of generality, we can assume sex≥y≥z. IMO 2000/2, Proposed by Titu Andreescu) Leta, b, cbe positive numbers such that abc= 1. The Ravi substitution is useful for inequalities about the lengths a, b, c of a triangle.
After Ravi's substitution, we can remove the condition that the lengths of the sides of the triangle. Tsintsifas) Let p, q, r be positive real numbers and let a, b, c denote the sides of a triangle with area F. V. Pambuccian) Using the Hadwiger-Finsler inequality, it suffices to show that p. However, this is a direct consequence of the Cauchy-Schwarz inequality. Note that this is a generalization of Weitzenb¨ock's inequality. (Why?) In [GC], G. Chang proved the Neuberg-Pedoe inequality using complex numbers.
Carlitz also observed that the Neuberg-Pedoe inequality can be deduced from Acz´el’s inequality. 0 and since the coefficient ofx2 in the quadratic polyno- mialP is positive,P should have at least one real root.
Trigonometric Methods
In fact, we get 1−4 sinBsinCcosBcosC = sin2B+ cos2B. in mathematics) transforms the problems in triangle geometry into trigonometric ones. Then we can use this to offer a lower bound for the perimeter of the triangle P QR:. sin 2Bsin 2C So one can consider the following inequality: Let r be the inradius of ABC and s=. On the one hand, we get aXA2+bXB2+cXC2.
Since the power of P with respect to the perimeter of ABC is OP2 =R2−BP ·P C, where R is the perimeter of the triangle ABC,. Let P be an interior point of a triangle ABC and let U, V, W be the points where the bisectors of the angles BP C, CP A, AP B intersect the sides BC, CA, AB respectively.
Applications of Complex Numbers
It is easy to check that the equality holds if and only if 4ABC is equilateral. Since the function f is not concave in R+, we cannot apply Jensen's inequality to the function f(t) = √1. In the first chapter, we discovered that the geometric inequality R≥2r is equivalent to the algebraic inequality abc ≥ (b+c−a)(c+a−b)(a+b−c).
In any triangle ABC, there is a natural relationship between cos A cosB+ cosC and Rr, where R and r are the radii of the circumcircle and circle of ABC.
Algebraic Substitutions
Note that the inequality is not symmetric in the three variables. 2For a verification of the identity, see [IV]. Here we give another solution of the problem 10. Latvia 2002) Leta,b,c,dbe the positive real numbers such that 1. given by Jeong Soo Sim at the KMO Weekend Program 2007).
Increasing Function Theorem
Establishing New Bounds
Long ago, an older and well-known number theorist made some disparaging remarks about the work of Paul Erdóos. You admire Erdos' contributions to mathematics as much as I do, and I felt irritated when the older mathematician stated bluntly and definitively that all of Erdos' work could be reduced to a few tricks that Erdos repeatedly relied on in his proofs. What the number theorist didn't realize is that other mathematicians, even the very best, also rely on a few tricks that they use over and over again.
It is sad to note that some of Hilbert's beautiful results have been completely forgotten. But reading the proofs of Hilbert's striking and profound theorems in invariant theory, it was surprising to confirm that Hilbert's proofs depended on the same few tricks.
Schur’s Inequality and Muirhead’s Theorem
Normalizations
After the substitution x= ab,y= bc,z= ac we obtain the constraint xyz= 1. From the constraint xyz= 1 we find two identities x. Applying the AM-GM inequality, we have p≥ 3. a) Determine the least constant C such that the inequality is X. b). Since the inequality is homogeneous, we can normalize tox1+ · · · +xn= 1. With x+y= 1 we homogenize the above inequality as follows.
Cauchy-Schwarz Inequality and H¨older’s Inequality
As an application of the Cauchy-Schwarz inequality, we give another solution to the following problem. If we use the same idea in the proof of the Cauchy-Schwarz inequality, we find a natural generalization. Modifying the above arguments slightly, we see that the AM-GM inequality implies that Theorem 3.4.5.
Recall that the AM-GM inequality is used to derive Theorem 18, which is a generalization of the Cauchy-Schwarz inequality. Now that we get the weighted version of the AM-GM inequality, we establish a weighted version of the Cauchy-Schwarz inequality. In the previous chapter, we derived the weighted AM-GM inequality from the AM-GM inequality.
We conclude this section by presenting a well-known inductive proof of the weighted Jensen's inequality.
Power Means
Using the convexity of xlnx or the convexity of xλ (λ ≥ 1), we can also determine the monotonicity of power means for nonpositive real numbers.
Majorization Inequality
Supporting Line Inequality
This means that the supporting line inequality is a powerful tool because we can also produce Jensen-type inequalities for non-convex functions. Because of the homogeneity of the inequality, we can normalize to a+b+c = 1. Every problem I solved became a rule that served to solve other problems afterwards. Determine the minimum C(n) for some values of n. where e the sums here and there are symmetric about the subscripts{1,· · · , n}.
Prove that for every integer n≥3 and every set of positive numbers x1, · · ·, xn, at least one of the two inequalities is satisfied. For each integer ≥2, find the largest constant Cn such that Cn. kak, where m is the largest integer not greater than cn. holds for every positive real number a.
Problems for Putnam Seminar
If x, y, and z are the radian measures of the angles of a (nondegenerate) triangle, prove that. with equality if and only ify= sinx. CMJ567 H.-J. Seiffert) Show that for all imaginary positive real numbers x and y, √. CMJ572, George Baloglou and Robert Underwood) Prove or disprove that for θ∈ −π4,π4. Prove that if A, B and C are the angles of an acute triangle and R is its circumradius.
IV Ilan Vardi, Solutions of the 2000 International Mathematical Olympiad http://www.lix.polytechnique.fr/Labo/Ilan.Vardi/publications.html.