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Algebra
4.9 The Rearrangements Inequality Problem 1.108 Let a, b, c be positive real numbers. Prove the inequality b c 3 a + + ≥ . b+c c+a a+b 2 Solution For symmetry reasons, we can assume with no loss of generality that a ≤ b ≤ c. But then 1 1 1 ≤ ≤ , b+c c+a a+b and the rearrangements inequality gives a b c b c a + + ≥ + + , b+c c+a a+b b+c c+a a+b and also b c c a b a + + ≥ + + . b+c c+a a+b b+c c+a a+b Adding the last two inequalities yields
b c a + + ≥ 3, 2 b+c c+a a+b which is what we wanted to prove. Problem 1.109 Let a, b, c be positive real numbers. Prove the inequality
b2
a3 b3 c3 a+b+c . + 2 + 2 ≥ 2 2 2 +c c +a a + b2
Solution Again, assume that a ≤ b ≤ c. This implies a 2 ≤ b2 ≤ c2 , and it is not difficult to check that b c a ≤ ≤ b 2 + c 2 c 2 + a 2 a 2 + b2 also holds. Applying the rearrangements inequality, we obtain
b2
a3 b3 c3 ab2 bc2 ca 2 + 2 + 2 ≥ 2 + 2 + 2 , 2 2 2 2 2 +c c +a a +b b +c c +a a + b2
and a3 b3 c3 ac2 ba 2 cb2 + + ≥ + + . b 2 + c 2 c 2 + a 2 a 2 + b2 b 2 + c 2 c 2 + a 2 a 2 + b 2