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Problem 2

Part of 2001 Mongolian Mathematical Olympiad

Problems(2)

inequality in triangle with medians

Source: Mongolia MO 2001 Grade 10 P2

4/12/2021
In an acute-angled triangle ABCABC, a,b,ca,b,c are sides, ma,mb,mcm_a,m_b,m_c the corresponding medians, RR the circumradius and rr the inradius. Prove the inequality a2+b2a+b⋅b2+c2b+c⋅a2+c2a+c≥16R2rmaa⋅mbb⋅mcc.\frac{a^2+b^2}{a+b}\cdot\frac{b^2+c^2}{b+c}\cdot\frac{a^2+c^2}{a+c}\ge16R^2r\frac{m_a}a\cdot\frac{m_b}b\cdot\frac{m_c}c.
geometric inequalityInequalitygeometry
inequality in two sequences

Source: Mongolia MO 2001 Teachers P2

4/12/2021
For positive real numbers b1,b2,…,bnb_1,b_2,\ldots,b_n define a1=b1b1+b2+…+bn and ak=b1+…+bkb1+…+bk−1 for k>1.a_1=\frac{b_1}{b_1+b_2+\ldots+b_n}\enspace\text{ and }\enspace a_k=\frac{b_1+\ldots+b_k}{b_1+\ldots+b_{k-1}}\text{ for }k>1.Prove that a1+a2+…+an≤1a1+1a2+…+1ana_1+a_2+\ldots+a_n\le\frac1{a_1}+\frac1{a_2}+\ldots+\frac1{a_n}
inequalitiesSequencealgebra