Problem: a , b , c a , b , c a 2 + b 2 + c 2 + ( a + b + c ) 2 ≤ 4 a 2 + b 2 + c 2 + ( a + b + c ) 2 ≤ 4
a b + 1 ( a + b ) 2 + b c + 1 ( b + c ) 2 + c a + 1 ( c + a ) 2 ≥ 3 ( a + b ) 2 a b + 1 ​ + ( b + c ) 2 b c + 1 ​ + ( c + a ) 2 c a + 1 ​ ≥ 3
Solution:
The given condition is equivalent to a 2 + b 2 + c 2 + a b + b c + c a ≤ 2 a 2 + b 2 + c 2 + a b + b c + c a ≤ 2
2 a b + 2 ( a + b ) 2 + 2 b c + 2 ( b + c ) 2 + 2 c a + 2 ( c + a ) 2 ≥ 6 ( a + b ) 2 2 a b + 2 ​ + ( b + c ) 2 2 b c + 2 ​ + ( c + a ) 2 2 c a + 2 ​ ≥ 6
Indeed, we have
2 a b + 2 ( a + b ) 2 ≥ 2 a b + a 2 + b 2 + c 2 + a b + b c + c a ( a + b ) 2 = 1 + ( c + a ) ( c + b ) ( a + b ) 2 ( a + b ) 2 2 a b + 2 ​ ≥ ( a + b ) 2 2 a b + a 2 + b 2 + c 2 + a b + b c + c a ​ = 1 + ( a + b ) 2 ( c + a ) ( c + b ) ​
Adding the last inequality with its cyclic analogous forms yields
2 a b + 2 ( a + b ) 2 + 2 b c + 2 ( b + c ) 2 + 2 c a + 2 ( c + a ) 2 ≥ 3 + ( c + a ) ( c + b ) ( a + b ) 2 + ( a + b ) ( a + c ) ( b + c ) 2 + ( b + c ) ( b + a ) ( c + a ) 2 ( a + b ) 2 2 a b + 2 ​ + ( b + c ) 2 2 b c + 2 ​ + ( c + a ) 2 2 c a + 2 ​ ≥ 3 + ( a + b ) 2 ( c + a ) ( c + b ) ​ + ( b + c ) 2 ( a + b ) ( a + c ) ​ + ( c + a ) 2 ( b + c ) ( b + a ) ​
Hence it remains to prove that
( c + a ) ( c + b ) ( a + b ) 2 + ( a + b ) ( a + c ) ( b + c ) 2 + ( b + c ) ( b + a ) ( c + a ) 2 ≥ 3 ( a + b ) 2 ( c + a ) ( c + b ) ​ + ( b + c ) 2 ( a + b ) ( a + c ) ​ + ( c + a ) 2 ( b + c ) ( b + a ) ​ ≥ 3
But this follows directly from the AM-GM inequality. Equality holds if and only if a + b = a + b = b + c = c + a b + c = c + a a = b = c = 1 3 a = b = c = 3 ​ 1 ​
OR
Set 2 x = a + b , 2 y = b + c 2 x = a + b , 2 y = b + c 2 z = c + a 2 z = c + a a = z + x − y , b = x + y − z a = z + x − y , b = x + y − z c = y + z − x c = y + z − x
a b + 1 ( a + b ) 2 = ( z + x − y ) ( x + y − z ) + 1 4 x 2 = x 2 − ( y − z ) 2 + 1 4 x 2 = x 2 + 2 y z + 1 − y 2 − z 2 4 x 2 ( a + b ) 2 a b + 1 ​ = 4 x 2 ( z + x − y ) ( x + y − z ) + 1 ​ = 4 x 2 x 2 − ( y − z ) 2 + 1 ​ = 4 x 2 x 2 + 2 y z + 1 − y 2 − z 2 ​
On the other hand, the given condition is equivalent to 2 a 2 + 2 b 2 + 2 c 2 + 2 a b + 2 b c + 2 c a ≤ 4 2 a 2 + 2 b 2 + 2 c 2 + 2 a b + 2 b c + 2 c a ≤ 4 ( a + b ) 2 + ( b + c ) 2 + ( c + a ) 2 ≤ 4 ( a + b ) 2 + ( b + c ) 2 + ( c + a ) 2 ≤ 4 x 2 + y 2 + z 2 ≤ 1 x 2 + y 2 + z 2 ≤ 1 1 − y 2 − z 2 ≥ x 2 1 − y 2 − z 2 ≥ x 2
a b + 1 ( a + b ) 2 = x 2 + 2 y z + 1 − y 2 − z 2 4 x 2 ≥ x 2 + 2 y z + x 2 4 x 2 = 1 2 + y z 2 x 2 ( a + b ) 2 a b + 1 ​ = 4 x 2 x 2 + 2 y z + 1 − y 2 − z 2 ​ ≥ 4 x 2 x 2 + 2 y z + x 2 ​ = 2 1 ​ + 2 x 2 y z ​
Likewise, we have
b c + 1 ( b + c ) 2 = 1 2 + z x 2 y 2 and c a + 1 ( c + a ) 2 = 1 2 + x y 2 z 2 ( b + c ) 2 b c + 1 ​ = 2 1 ​ + 2 y 2 z x ​ and ( c + a ) 2 c a + 1 ​ = 2 1 ​ + 2 z 2 x y ​
Adding the last three inequalities gives
a b + 1 ( a + b ) 2 + b c + 1 ( b + c ) 2 + c a + 1 ( c + a ) 2 ≥ 3 2 + y z 2 x 2 + z x 2 y 2 + x y 2 z 2 ≥ 3 ( a + b ) 2 a b + 1 ​ + ( b + c ) 2 b c + 1 ​ + ( c + a ) 2 c a + 1 ​ ≥ 2 3 ​ + 2 x 2 y z ​ + 2 y 2 z x ​ + 2 z 2 x y ​ ≥ 3
by the AM-GM inequality. Equality holds if and only if x = y = z x = y = z a = b = c = 1 3 a = b = c = 3 ​ 1 ​
The problems on this page are the property of the MAA's American Mathematics Competitions