Advanced book on Mathematics Olympiad

Algebra 387

Theorem.Ifa 1 ,a 2 ,a 3 ,b 1 ,b 2 ,b 3 are real numbers such that

a 1 ≥a 2 ≥a 3 ≥ 0 ,b 1 ≥b 2 ≥b 3 ≥ 0 ,a 1 ≥b 1 ,a 1 +a 2 ≥b 1 +b 2 ,

a 1 +a 2 +a 3 =b 1 +b 2 +b 3 ,

then for any positive real numbersx, y, z, one has

∑

sym

xa^1 ya^2 za^3 ≥

∑

sym

xb^1 yb^2 zb^3 ,

where the indexsymsignifies that the summation is over all permutations ofx, y, z.

Using the fact thatabc=1, we rewrite the inequality as

1

a^3 (b+c)

+

1

b^3 (c+a)

+

1

c^3 (a+b)

≥

3

2 (abc)^4 /^3

.

Seta=x^3 ,b=y^3 ,c=z^3 , withx, y, z >0. The inequality becomes

∑

cyclic

1

x^9 (y^3 +z^3 )

≥

3

2 x^4 y^4 z^4

.

Clearing denominators, this becomes

∑

sym

x^12 y^12 + 2

∑

sym

x^12 y^9 z^3 +

∑

sym

x^9 y^9 z^6 ≥ 3

∑

sym

x^11 y^8 z^5 + 6 x^8 y^8 z^8 ,

or

(

∑

sym

x^12 y^12 −

∑

sym

x^11 y^8 z^5

)

+ 2

(

∑

sym

x^12 y^9 z^3 −

∑

sym

x^11 y^8 z^5

)

+

(

∑

sym

x^9 y^9 z^6 −

∑

sym

x^8 y^8 z^8

)

≥ 0.

And every term on the left-hand side is nonnegative by the Muirhead inequality.
(36th International Mathematical Olympiad, 1995)
145.ViewQas a polynomial inx. It is easy to see thatyis a zero of this polynomial;
henceQis divisible byx−y. By symmetry, it is also divisible byy−zandz−x.
146.The relation(x+ 1 )P (x)=(x− 10 )P (x+ 1 )shows thatP(x)is divisible by(x− 10 ).
Shifting the variable, we obtain the equivalent relationxP(x− 1 )=(x− 11 )P (x),
which shows thatP(x)is also divisible byx. HenceP(x)=x(x− 10 )P 1 (x)for some
polynomialP 1 (x). Substituting in the original equation and canceling common factors,
we find thatP 1 (x)satisfies