366 Algebra
So a good candidate for the minimum is 2n, which is actually attained forx 1 =x 2 =
··· =xn=^12.
(Romanian Mathematical Olympiad, 1984, proposed by T. Andreescu)
99.Assume the contrary, namely that 7a+ 5 b+ 12 ab >9. Then
9 a^2 + 8 ab+ 7 b^2 −( 7 a+ 5 b+ 12 ab) < 6 − 9.Hence
2 a^2 − 4 ab+ 2 b^2 + 7(
a^2 −a+1
4
)
+ 5
(
b^2 −b+1
4
)
< 0 ,
or
2 (a−b)^2 + 7(
a−1
2
) 2
+ 5
(
b−1
2
) 2
< 0 ,
a contradiction. The conclusion follows.
(T. Andreescu)
100.We rewrite the inequalities to be proved as− 1 ≤ak−n≤1. In this respect,
we have
∑nk= 1(ak−n)^2 =∑nk= 1a^2 k− 2 n∑nk= 1ak+n·n^2 ≤n^3 + 1 − 2 n·n^2 +n^3 = 1 ,and the conclusion follows.
(Math Horizons, proposed by T. Andreescu)
101.Adding up the two equations yields
(
x^4 + 2 x^3 −x+
1
4
)
+
(
y^4 + 2 y^3 −y+1
4
)
= 0.
Here we recognize two perfect squares, and write this as
(
x^2 +x−1
2
) 2
+
(
y^2 +y−1
2
) 2
= 0.
Equality can hold only ifx^2 +x−^12 =y^2 +y−^12 =0, which then gives{x, y}⊂
{−^12 −
√ 3
2 ,−1
2 +√ 3
2 }. Moreover, sincex =y,{x, y}={−1
2 −√ 3
2 ,−1
2 +√ 3
2 }. A simple
verification leads to(x, y)=(−^12 +
√
3
2 ,−1
2 −√
3
2 ).
(Mathematical Reflections, proposed by T. Andreescu)102.Letn= 2 k. It suffices to prove that