Advanced book on Mathematics Olympiad

384 Algebra

Subtractingn(n 2 +^1 )from this expression, we obtain

na^3 + 3

[

n(n+ 1 )

2 −k

]

a^2 +

[

n(n+ 1 )( 2 n+ 1 )

2 −^3 k

(^2) −n^2 (n+^1 )
2

]

a−k^3 +n(n 2 +^1 )k

na+n(n 2 +^1 )−k

.

The numerator is the smallest whenk=nanda=1, in which case it is equal to 0.
Otherwise, it is strictly positive, proving that the minimum is not attained in that case.
Therefore, the desired minimum isn(n 2 +^1 ), attained only ifxk=k,k= 1 , 2 ,...,n.
(American Mathematical Monthly, proposed by C. Popescu)

139.First, note that the inequality is obvious if eitherxoryis at least 1. For the case
x, y∈( 0 , 1 ), we rely on the inequality

ab≥

a

a+b−ab

,

which holds fora, b∈( 0 , 1 ). To prove this new inequality, write it as

a^1 −b≤a+b−ab,

and then use the Bernoulli inequality to write

a^1 −b=( 1 +a− 1 )^1 −b≤ 1 +(a− 1 )( 1 −b)=a+b−ab.

Using this, we have

xy+yx≥

x

x+y−xy

+

y

x+y−xy

>

x

x+y

+

y

x+y

= 1 ,

completing the solution to the problem,
(French Mathematical Olympiad, 1996)

140.We have

x^5 −x^2 + 3 ≥x^3 + 2 ,

for allx≥0, because this is equivalent to(x^3 − 1 )(x^2 − 1 )≥0. Thus

(a^5 −a^2 + 3 )(b^5 −b^2 + 3 )(c^5 −c^2 + 3 )≥(a^3 + 1 + 1 )( 1 +b^3 + 1 )( 1 + 1 +c^3 ).

Let us recall Hölder’s inequality, which in its most general form states that forr 1 ,r 2 ,...,
rk>0, withr^11 +r^12 +···+r^1 k=1 and for positive real numbersaij,i= 1 , 2 ,...,k,
j= 1 , 2 ,...,n,

∑n

i= 1

a 1 ia 2 i···aki≤

( n

∑

i= 1

ar 11 i

)r^1

1

(n

∑

i= 1

a 2 r^2 i

)r^1

2

···

(n

∑

i= 1

arkik

)r^1

k

.