Basics of olympiad inequalities by Riasat S. By Riasat S.

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3. 4. Solution: WLOG assume that a ≥ b ≥ c. This implies a + b ≥ c + a ≥ b + c. Therefore a b c ≥ ≥ b+c c+a a+b On the other hand, we have ab ≥ ca ≥ bc. Therefore a(b + c) ≥ b(c + a) ≥ c(a + b). Applying Chebyshev’s a b c inequality for the similarly sorted sequences b+c , c+a , a+b and (a(b + c), b(c + a), c(a + b)) we get 3 a · a(b + c) ≥ b+c a b+c ⇔ 3(a2 + b2 + c2 ) ≥ 2(ab + bc + ca) which was what we wanted. a(b + c) a b+c 36 CHAPTER 6. 5. Solution: Let x = an b+c , y √ n But the sequences ( an−1 x, = n bn c+a , z bn−1 y, = cn a+b .

Let a = x + y, b = y + z and c = z + x. 3. Second Solution. 5. NORMALIZATION 29 which is Schur’s inequality. 2. Let a, b, c be the lengths of the sides of a triangle. Prove that √ 3 √ ab + bc + √ ca ≥ √ a+b−c+ √ b+c−a+ √ c + a − b. Solution. Let x, y, z > 0 such that a = x + y, b = y + z, c = z + x. Then our inequality is equivalent to √ (x + y)(y + z) ≥ 2 3 cyc 2 x . cyc From Cauchy-Schwarz inequality, (x + y)(y + z) ≥ 3 3 cyc (y + √ zx) cyc ≥2 √ y+4 cyc √ =2 zx cyc 2 x . g. applied restrictions with homogeneous expressions in the variables.

This implies a b c = 1, and our inequality becomes a 3 + b 3 + c 3 − 3a b c ≥ 0, which is the same as before. Therefore the restriction abc = 1 doesn’t change anything of the inequality. Similarly one might also assume a + b + c = 1. The reader is requested to find out how it works. 6 Homogenization This is the opposite of Normalization. It is often useful to substitute a = x/y, b = y/z, c = z/x, when the condition abc = 1 is given. Similarly when a + b + c = 1 we can substitute a = x/x + y + z, b = y/x + y + z, c = z/x + y + z to homogenize the inequality.