Inequalities (Unit 1 â Unit 3) Solutions to Exercises Inequalities (Unit ...

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222Equality holds if and only if x1+ x2= 2x1x2, or ( x1− x2) = 0 , i.e. x1= x2.Hence equality of the original inequality holds if and only if any (n – 2) of the x i’s arezero and the remaining two xi’s are equal (possibly to zero).10
Inequalities (Unit 3)1. By Heron’s formula,T⎛a+ b+ c⎞⎛− a+ b+ c⎞⎛a− b+ c⎞⎛a+ b−c⎞= ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ 2 ⎠⎝ 2 ⎠⎝ 2 ⎠⎝ 2 ⎠ .Putting this into the original inequality, the original inequality can be simplified as follows:2 2 2a + b + c ≥ 3( a+ b+ c)( − a+ b+ c)( a− b+ c)( a+ b−c)4 4 4 2 2 2 2 2 2 2 2 2 2a + b + c + 2a b + 2b c + 2c a ≥ 3 ⎡⎣( b+ c) −a ⎤⎡ ⎦⎣a −( b−c)⎤⎦4 4 4 2 2 2 2 2 2 2 2 2 2 2 2a + b + c + 2a b + 2b c + 2c a ≥ 3⎡⎣ 2 bc+ ( b + c −a ) ⎤⎡ ⎦⎣2 bc− ( b + c −a) ⎤⎦4 4 4 2 2 2 2 2 2 2 2 2 2 2 2 4 4 4a + b + c + 2ab + 2bc + 2ca ≥ 3[ 2ab + 2bc + 2ca −a −b −c]4 4 4 2 2 2 2 2 24a + 4b + 4c ≥ 4a b + 4b c + 4c aBy the AM-GM inequality, we have4 4 4 4 4 4 4 4 44a + 4b + 4 c = (2a + 2 b ) + (2b + 2 c ) + (2c + 2 a )4 4 4 4 4 4( ) ( ) ( )≥ 2 2a ⋅ 2b + 2 2b ⋅ 2c + 2 2c ⋅2a2 2 2 2 2 2= 4ab + 4bc + 4cathereby proving the last inequality and hence the original inequality. It is clear in theapplication of the AM-GM inequality that equality holds if and only if a = b= c.Alternative SolutionWithout loss of generality, assume that the angle opposite the side a is acute. Suppose that thealtitude from this vertex, whose length we denote by h, is of distances m and n from theremaining 2 vertices, with b= 2 2h + m and c= 2 2h + n . WLOG, assume that m ≥ n. Thena=m+n or m-n.For a=m+n, the original inequality becomes2 2 2 2 2 ( m+n)h( m+ n) + ( h + m ) + ( h + n ) ≥4 3⋅ .2Rewriting this as a quadratic equality in h, we have2 2 2h − 3( m+ n) h+ ( m + mn+ n ) ≥ 0.The discriminant of the quadratic function on the left is22 2 23( m n) ⎤ 4(1)( m mn n ) ( m n) 0Δ = ⎣⎡+ ⎦ − + + =− − ≤ .Since the coefficient of h 2 is positive, this means h 2 − 3( m+ n) h+ ( m 2 + mn+ n2 ) ≤ 0 for allh, as desired. Equality holds when m= n , which means b= c . By symmetry, we needa = b= c.For a=m-n, the argument is the same as above, with n replaced by –n.11
Page 2 and 3: By the AM-GM inequality, we haveSim
Page 5 and 6: xy + yz + zx − 3 xyz = xy(1 − z
Page 7: n∑k = 12⎛ a⎜k⎝ ak+ bk⎞⎟
Page 13 and 14: Hence the AM-GM inequality asserts
Page 15: andRRCE1 ⎛csin A bsin B esin B f

Inequalities (Unit 1 â Unit 3) Solutions to Exercises Inequalities (Unit ...

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Inequalities (Unit 1 â Unit 3) Solutions to Exercises Inequalities (Unit ...