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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 a+b+c \le
2abc+\sqrt2.

(Proposed by János Bodnár, Budapest)

(5 pont)

Deadline expired on October 15, 2007.


Solution 1. Let u=1-\sqrt2a, v=1-\sqrt2b and w=1-\sqrt2c. 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)
 uv+vw+wu = \frac12(u+v+w-1)^2. (1c)

The statement to prove is

 (1-u)+(1-v)+(1-w) = \sqrt2(a+b+c) \le 2\sqrt2abc+2 = (1-u)(1-v)(1-w)+2;

uvw\leuv+vw+wu.

Case 1: u,v,w\ge0. From the constraint |1-u|\le\sqrt2, hence u\le\sqrt2+1<3; it can be obtained similarly that v<3 and w<3. Therefore

 uv+vw+wu \ge 
uv\cdot\frac{w}3 +
vw\cdot\frac{u}3 +
wu\cdot\frac{v}3 = uvw.

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

 uvw \le 0 \le \frac12(u+v+w-1)^2 = uv+vw+wu.

Solution 2. By the constraint we have 2ab\lea2+b2\lea2+b2+c2=1 and similarly 2ac\le1, 2bc\le1.

Then

2-(a+b+c-2abc)2=1+(a2+b2+c2)-(a+b+c-2abc)2=(1-2ab)(1-2ac)(1-2bc)+4a2b2c2\ge0

and therefore

 |a+b+c-2abc|\le\sqrt2.

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.

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