Mathematical and Physical Journal
for High Schools
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# Problem A. 433. (September 2007)

A. 433. Prove that if a, b, c are real numbers such that a2+b2+c2=1 then .

(Proposed by János Bodnár, Budapest)

(5 pont)

Deadline expired on October 15, 2007.

Solution 1. Let , and . For these variables the constraint can be re-written as

 (1-u)2+(1-v)2+(1-w)2=2(a2+b2+c2)=2; (1a)
 u2+v2+w2=2(u+v+w)-1; (1b)
 (1c)

The statement to prove is

uvwuv+vw+wu.

Case 1: u,v,w0. From the constraint , hence ; it can be obtained similarly that v<3 and w<3. Therefore

Case 2: at least one of u,v,w is negative. Equation (1a) cannot hold if two of them are negative; so exactly one of u,v,w is negative. Then

Solution 2. By the constraint we have 2aba2+b2a2+b2+c2=1 and similarly 2ac1, 2bc1.

Then

2-(a+b+c-2abc)2=1+(a2+b2+c2)-(a+b+c-2abc)2=(1-2ab)(1-2ac)(1-2bc)+4a2b2c20

and therefore

From the solution of Tuan Nhat Le

### Statistics:

 16 students sent a solution. 5 points: Blázsik Zoltán, Huszár Kristóf, Korándi Dániel, Lovász László Miklós, Nagy 235 János, Nagy 314 Dániel, Szűcs Gergely, Tomon István, Tossenberger Anna, Tuan Nhat Le, Wolosz János. 1 point: 1 student. 0 point: 4 students.

Problems in Mathematics of KöMaL, September 2007