Advanced book on Mathematics Olympiad

374 Algebra

120.We start with the algebraic identity

x^3 (y−z)+y^3 (z−x)+z^3 (x−y)=(x+y+z)(x−y)(y−z)(z−x),

wherex, y, zare complex numbers. Applying to it the triangle inequality, we obtain

|x|^3 |y−z|+|y|^3 |z−x|+|z|^3 |x−y|≥|x+y+z||x−y||x−z||y−z|.

So let us see how this can be applied to our problem. Place the triangle in the complex

plane so thatMis the origin, and leta,b, andc, respectively, be the complex coordinates

ofA,B,C. The coordinate ofGis(a+b 3 +c), and if we setx=a,y=b, andz=cin the

inequality we just derived, we obtain the geometric inequality from the statement.

(M. Dinc ̆a, M. Chiri ̧ta, ̆Numere Complexe în Matematica de Liceu(Complex Numbers

in High School Mathematics), ALL Educational, Bucharest, 1996)

121.BecauseP(x)has odd degree, it has a real zeror.Ifr>0, then by the AM–GM

inequality

P(r)=r^5 + 1 + 1 + 1 + 25 − 5 · 2 ·r≥ 0.

And the inequality is strict since 1 =2. Hencer<0, as desired.

122.We can rewrite the inequality as

nn− 1

n− 1

≥n

n+ 21

,

or

nn−^1 +nn−^2 +···+ 1 ≥n

n+ 21

.

This form suggests the use of the AM–GM inequality, and indeed, we have

1 +n+n^2 +···+nn−^1 ≥nn

√

1 ·n·n^2 ···nn−^1 =n

n

√

n

n(n 2 − 1 )

=n

n+ 21

,

which proves the inequality.

(Gh. Calug ̆ ari ̧ta, V. Mangu, ̆ Probleme de Matematic ̆a pentru TreaptaI ̧si aII-ade

Liceu(Mathematics Problems for High School), Editura Albatros, Bucharest, 1977)

123.The inequality is homogeneous in the sense that if we multiply someakandbk

simultaneously by a positive number, the inequality does not change. Hence we can

assume thatak+bk=1,k= 1 , 2 ,...,n. In this case, applying the AM–GM inequality,

we obtain

(a 1 a 2 ···an)^1 /n+(b 1 b 2 ···bn)^1 /n≤

a 1 +a 2 +···+an

n

+

b 1 +b 2 +···+bn

n

=

a 1 +b 1 +a 2 +b 2 +···+an+bn

n

=

n

n

= 1 ,