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Deduce a proof of Nesbitt’s inequality from Chebyshev’s inequality. 4. Let a1 ≥ a2 ≥ · · · ≥ an and b1 ≤ b2 ≤ · · · ≤ bn be positive. Prove that n i=1 ai a1 + a2 + · · · + an n(a1 + a2 + · · · + an ) √ ≥ . ≥ n bi b1 + b2 + · · · + bn b1 b2 . . 5. Let a ≥ c ≥ 0 and b ≥ d ≥ 0. Prove that (a + b + c + d)2 ≥ 8(ad + bc). 6. (Radu Titiu) Let a, b, c > 0 such that a2 + b2 + c2 ≥ 3. Show that a2 b2 c2 3 + + ≥ . 3. 7. Let a, b, c > 0 such that abc = 1. Prove that 9 ≤ 2 cyc 1 · + b) c2 (a ab ≤ 3 cyc cyc ab .
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. 1 in the following form: a b c a+b+c + + ≥ √ . 3 b c a abc On the other hand, if we substitute a = x/y, b = y/z, c = z/x the inequality becomes, zx xy yz x y z + 2 + 2 ≥ + + , 2 y z x y z x which clearly looks easier to deal with (Hint: Rearrangement).
B+c bn + bn−1 c c+a cyc cyc cyc 26 CHAPTER 3. 1 Schur’s Inequality Let a, b, c be positive real numbers, and n be positive. Then the following inequality holds: an (a − b)(a − c) + bn (b − c)(b − a) + cn (c − a)(c − b) ≥ 0, with equality if and only if a = b = c or a = b, c = 0 and permutations. The above inequality is known as Schur’s inequality, after Issai Schur. Proof. Since the inequality is symmetric in a, b, c WLOG we may assume that a ≥ b ≥ c. Then the inequality is equivalent to (a − b)(an (a − c) − bn (b − c)) + cn (a − c)(b − c) ≥ 0, which is obviously true.