Advanced book on Mathematics Olympiad

376 Algebra

126.The solution is based on the Lagrange identity, which in our case states that ifMis

a point in space andGis the centroid of the tetrahedronABCD, then

AB^2 +AC^2 +CD^2 +AD^2 +BC^2 +BD^2

= 4 (MA^2 +MB^2 +MC^2 +MD^2 )− 16 MG^2.

ForM=Othe center of the circumscribed sphere, this reads

AB^2 +AC^2 +CD^2 +AD^2 +BC^2 +BD^2 = 16 − 16 OG^2.

Applying the AM–GM inequality, we obtain

63

√

AB·AC·CD·AD·BC·BD≤ 16 − 16 OG^2.

This combined with the hypothesis yields 16≤ 16 −OG^2. So on the one hand we

have equality in the AM–GM inequality, and on the other handO=G. Therefore,

AB=AC=AD=BC=BD=CD, so the tetrahedron is regular.

127.Adding 1 to all fractions transforms the inequality into

x^2 +y^2 + 1

2 x^2 + 1

+

y^2 +z^2 + 1

2 y^2 + 1

+

z^2 +x^2 + 1

2 z^2 + 1

≥ 3.

Applying the AM–GM inequality to the left-hand side gives

x^2 +y^2 + 1

2 x^2 + 1

+

y^2 +z^2 + 1

2 y^2 + 1

+

z^2 +x^2 + 1

2 z^2 + 1

≥ 33

√

x^2 +y^2 + 1

2 x^2 + 1

·

y^2 +z^2 + 1

2 y^2 + 1

·

z^2 +x^2 + 1

2 z^2 + 1

.

We are left with the simpler but sharper inequality

x^2 +y^2 + 1

2 x^2 + 1

·

y^2 +z^2 + 1

2 y^2 + 1

·

z^2 +x^2 + 1

2 z^2 + 1

≥ 1.

This can be proved by multiplying together

x^2 +y^2 + 1 =x^2 +

1

2

+y^2 +

1

2

≥ 2

√(

x^2 +

1

2

)(

y^2 +

1

2

)

,

y^2 +z^2 + 1 =y^2 +

1

2

+z^2 +

1

2

≥ 2

√(

y^2 +

1

2

)(

z^2 +

1

2

)

,

z^2 +x^2 + 1 =z^2 +

1

2

+x^2 +

1

2

≥ 2

√(

z^2 +

1

2

)(

y^2 +

1

2

)

,