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USING COMPLEX NUMBER TO PROVE INEQUALITIES Batigoal_mathscope.org Email: hoangquan9@gmail.com I.Theorem Let a, b, a ', b ' be real numbers Let complex numbers z = a + bi and z’ = a '+ b ' i ( i = −1 ) We have z + z ' ≥ z + z ' II.Application Example 1: Let x, y be real numbers Prove that x − x + + x + x + 17 ≥ 29 Solution We have x − x + + x + x + 17 ≥ 29 ⇔ ( x − 1) + + ( x + 1) + 16 ≥ 29 Let complex numbers z = x − + i , z ' = − x − + 4i and z '' = −2 + 5i We have: z + z ' = z '' using inequality z + z ' ≥ z + z ' , we have ( x − 1)2 + + ( x + 1) + 16 ≥ 29 Example 2: Let a1 , a2 , b1 , b2 be real numbers Prove that: (a1 + a2 ) + (b1 + b2 )2 ≤ a12 + b12 + a22 + b22 Solution Let complex numbers z = a1 + b1i and z ' = a2 + b2i We have z + z ' = (a1 + a2 ) + (b1 + b2 )i using inequality z + z ' ≥ z + z ' ,we have: (a1 + a2 ) + (b1 + b2 ) ≤ a12 + b12 + a22 + b22 Example 3: Let a, b, c be real numbers Prove that: a + ab + b + a + ac + c ≥ b + bc + c Solution We have a + ab + b + a + ac + c ≥ b + bc + c 2 2 2 b b 3 c c 3 b c b c 3 ⇔ a + + + a + + ≥ − + + 2 b Let complex numbers z = a + + b c c i , z ' = −a − + i 2 2 c b b c 3 We have z + z ' = − + + i 2 2 using inequality z + z ' ≥ z + z ' , we have: 2 2 2 b b 3 c c 3 b c b c 3 a + + + a + + ≥ − + + 2 Example 4: Let x, y, z be positive real numbers such that x + y + z = Prove that: x + xy + y + y + yz + z + x + xz + z ≥ 3 Solution Let S = x + xy + y + y + yz + z + x + xz + z 2 3y y z 3z x 3x We have S = ( x + ) + + ( y + ) + + ( z + ) + 2 y 3y z 3z i , z ' = y + + i , Let complex numbers z = x + + z '' = z + x 3x + i 3 ( x + y + z)i 2 using inequality z + z ' + z '' ≥ z + z '+ z '' , we have: We have z + z '+ z '' = ( x + y + z ) + 2 3y y z 3z x 3x ( x + ) + ( x + y + z )2 + ( x + y + z ) + ( y + ) + + ( z + ) + ≥ 2 4 2 3y y z 3z x 3x 9 + = 27 ⇔ ( x + )2 + + ( y + ) + + ( z + ) + ≥ 2 4 Thus 2 3y y z 3z x 3x ( x + ) + + ( y + ) + + ( z + ) + ≥ 3 2 Example5: Let a, b, c be positive real numbers such that ab + bc +ca = abc Prove that: a + 2b b + 2c c + 2a + + ≥ ab bc ca Solution Let x = 1 , y = , z = , we have: x, y , z > and x + y + z =1 a b c a + 2b b + 2c c + 2a LHS = = x2 + y + y2 + z + z + x2 + + ab bc ca Let complex numbers z = x + yi , z ' = y + zi , z '' = z + xi We have z + z '+ z '' = ( x + y + z ) + 2( x + y + z )i using inequality z + z ' + z '' ≥ z + z '+ z '' , we have: x2 + y + y + z + z + x2 ≥ (x + y + z) + 2( x + y + z) ⇔ x + y + y + z + z + x ≥ (because x + y + z = 1) Thus a + 2b b + 2c c + 2a + + ≥ ab bc ca Example6: Let a, b, c be positive real numbers such that x + y + z ≤ Prove that x2 + 1 + y + + z + ≥ 82 x y z Solution x y z Let complex numbers z = x + i , z ' = y + i , z '' = z + i 1 We have z + z '+ z '' = ( x + y + z ) + ( + + )i x y z using inenquality z + z ' + z '' ≥ z + z '+ z '' , we have: 1 1 1 + y + + z + ≥ ( x + y + z )2 + ( + + ) 2 x y z x y z On the other hand, we have 1 1 1 ( x + y + z ) + ( + + )2 = 81( x + y + z ) + ( + + )2 − 80( x + y + z ) x y z x y z We will use the AM-GM inequality, we have 1 1 1 81( x + y + z ) + ( + + ) ≥ 81( x + y + z ) ( + + ) ≥ 18.9 x y z x y z S = x2 + thus 1 1 1 ( x + y + z ) + ( + + ) = 81( x + y + z )2 + ( + + ) − 80( x + y + z )2 ≥ 18.9 − 80.1 = 82 x y z x y z 1 then LHS = x + + y + + z + ≥ 82 x y z Batigoal_mathscope.org Email: hoangquan9@gmail.com ... positive real numbers such that x + y + z = Prove that: x + xy + y + y + yz + z + x + xz + z ≥ 3 Solution Let S = x + xy + y + y + yz + z + x + xz + z 2 3y y z 3z x 3x We have S = (... positive real numbers such that ab + bc +ca = abc Prove that: a + 2b b + 2c c + 2a + + ≥ ab bc ca Solution Let x = 1 , y = , z = , we have: x, y , z > and x + y + z =1 a b c a + 2b b + 2c c + 2a... a, b, c be positive real numbers such that x + y + z ≤ Prove that x2 + 1 + y + + z + ≥ 82 x y z Solution x y z Let complex numbers z = x + i , z ' = y + i , z '' = z + i 1 We have z + z '+ z ''
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