Maths Olympiad Prep

Library / Turkey — Team Selection Test /17 of 66

Turkey — Team Selection Test

Algebra Difficulty 7.8 National Olympiad, round 2 Prove it Turkey

Find the greatest constant MM such that
a2+b2+c2+3abc≥M(ab+bc+ca) a^2 + b^2 + c^2 + 3abc \geq M(ab + bc + ca)
for all nonnegative real numbers a,b,ca, b, c satisfying a+b+c=4a + b + c = 4.

Solution

Letting a=0a = 0 and b=c=2b = c = 2 we obtain 2≥M2 \ge M. We will show that M=2M = 2 works.

Without loss of generality we may assume that max⁡{a,b,c}=c\max\{a, b, c\} = c. Let x=a+bx = a + b and y=aby = ab. We have c≥a+b+c3=43c \ge \frac{a+b+c}{3} = \frac{4}{3} and hence x=a+b≤83x = a+b \le \frac{8}{3}.

Then
a2+b2+c2+3abc≥2(ab+bc+ca)  ⟺  x2−2y+(4−x)2+3y(4−x)≥2(y+x(4−x))  ⟺  4(x−2)2+y(8−3x)≥0 a^2 + b^2 + c^2 + 3abc \ge 2(ab + bc + ca) \iff x^2 - 2y + (4-x)^2 + 3y(4-x) \ge 2(y+x(4-x)) \iff 4(x-2)^2 + y(8-3x) \ge 0
follows.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.