Advanced book on Mathematics Olympiad

Algebra 379

This implies

1

8

−

1

4

+

1

2

(ab+bc+ca)−abc < 0 ,

and so 4(ab+bc+ca)− 8 abc≤1, and the desired inequality holds.

If^12 −a≥0,^12 −b≥0,^12 −c≥0, then

2

√(

1

2

−a

)(

1

2

−b

)

≤

(

1

2

−a

)

+

(

1

2

−b

)

= 1 −a−b=c.

Similarly,

2

√(

1

2

−b

)(

1

2

−c

)

≤a and 2

√(

1

2

−c

)(

1

2

−a

)

≤b.

It follows that

8

(

1

2

−a

)(

1

2

−b

)(

1

2

−c

)

≤abc,

and the desired inequality follows.

(Mathematical Reflections, proposed by T. Andreescu)

131.Ifxi <xj for somei andj, increasexiand decreasexj by some numbera,

0 <a≤xj−xi. We need to show that

(

1 +

1

xi+a

)(

1 +

1

xj−a

)

<

(

1 +

1

xi

)(

1 +

1

xj

)

,

or

(xi+a+ 1 )(xj−a+ 1 )

(xi+a)(xj−a)

<

(xi+ 1 )(xj+ 1 )

xixj

.

All denominators are positive, so after multiplying out and canceling terms, we obtain
the equivalent inequality

−axi^2 +axj^2 −a^2 xi−a^2 xj−axi+axj−a^2 > 0.

This can be rewritten as

a(xj−xi)(xj+xi+ 1 )>a^2 (xj+xi+ 1 ),

which is true, sincea<xj−xi. Starting with the smallest and the largest of the numbers,

we apply the trick and make one of the numbers equal to^1 nby decreasing the value of