Advanced book on Mathematics Olympiad

370 Algebra

112.Bring the denominator to the left:

(a+b)(b+c)(c+a)

(

1

a+b

+

1

b+c

+

1

c+a

+

1

23

√

abc

)

≥(a+b+c+^3

√

abc)^2.

The identity

(a+b)(b+c)(c+a)=c^2 (a+b)+b^2 (c+a)+a^2 (b+c)+ 2 abc

enables us to transform this into

(c^2 (a+b)+b^2 (c+a)+a^2 (b+c)+ 2 abc)

(

1

a+b

+

1

b+c

+

1

c+a

+

1

23

√

abc

)

≥(c+b+a+^3

√

abc)^2.

And now we recognize the Cauchy–Schwarz inequality. Equality holds only ifa=b=c.
(Mathematical Olympiad Summer Program, T. Andreescu)
113.Letcbe the largest side. By the triangle inequality,cn<an+bnfor alln≥1. This
is equivalent to

1 <

(a

c

)n

+

(

b

c

)n

,n≥ 1.

Ifa<candb<c, then by lettingn→∞, we obtain 1<0, impossible. Hence one of

the other two sides equalsc, and the triangle is isosceles.

114.Defined=−a−b−c. The inequality becomes

‖a‖+‖b‖+‖c‖+‖d‖≥‖a+d‖+‖b+d‖+‖c+d‖.

If the angles formed byawithb,c, anddcome in increasing order, then the closed

polygonal linea,b,c,dis a convex quadrilateral. Figure 59 shows how this quadrilateral

can be transformed into one that is skew by choosing one angle such that one of the pairs

of adjacent angles containing it totals at most 180◦and the other at least 180◦and then

folding that angle in.

The triangle inequality implies‖b‖+‖c‖≥‖b+d‖+‖c+d‖. To be more convincing,

let us explain that the left-hand side is the sum of the lengths of the dotted segments,

while the right-hand side can be decomposed into the lengths of some four segments,

which together with the dotted segments form two triangles. The triangle inequality also

gives‖a‖+‖d‖≥‖a+d‖. Adding the two yields the inequality from the statement.

(Kvant(Quantum))

115.Letλ 1 ,λ 2 ,...,λnbe the roots of the polynomial,D 1 ={z, |z−c|≤R}the

disk covering them, andD 2 ={z,|z−c|≤R+|k|}. We will show that the roots of

nP (z)−kP′(z)lie insideD 2.