Suppose real numbers A,B,C such that for all real numbers x,y,z the following inequality holds: A(x−y)(x−z)+B(y−z)(y−x)+C(z−x)(z−y)≥0. Find the necessary and sufficient condition A,B,C must satisfy (expressed by means of an equality or an inequality).
Problem
Source: China TST 1988, problem 1
Tags: inequalities, quadratics, inequalities unsolved
28.06.2005 11:46
very not sure about my "proof" but thought i would post anyway A(x−y)(x−z)+B(y−z)(y−x)+C(z−x)(z−y)≥0. (*) Suppose (*). Without loss of generality, x≥y≥z. [We can repeat the same argument if x≥z≥y, etc.] Let a=x−y and b=y−z and understand that a,b are positive reals. Then: (∗)⇔A(x−y)(x−y+y−z)+C(x−y+y−z)(y−z)≥B(x−y)(y−z) ⇔Aa(a+b)+C(a+b)b≥Bab ⇔A(ab)2+(A+C−B)(ab)+C≥0 (If b=0, the condition would be just A≥0). Notice that ab≥0. If A≤0, for extreme values of a/b, we have it being negative. So certainly A>0. If there is a positive root, then some segment of the graph of the quadratic is in the 4th quadrant. So we want −(A+C−B)+√A2+B2+C2−2AB−2BC−2CA<0, or A+C−B>√(A+C−B)2−4AC, which is true if AC>0. From here, using AMGM it is clear that the condition is 4√AC≥B. Because our steps are reversible, it is clear it is sufficent.
14.10.2016 05:00
I think your proof is wrong! There you can't suppose x>y>z cause it surely loses generality. Remember the inequality is a family of inequalities about A,B,C ,you can't just change the condition of the inequities. My proof is to suppose z=0,the use discriminant of a quadratic equation to get that the square root of A,B,C forms a triangle.
14.10.2016 05:10
I think that you can suppose x>y>z. If you do this, depending on if it's true or not, you can just change A, B, and C to fit the inequality. Also, It would be nice if you could elaborate on how ⇔Aa(a+b)+C(a+b)b≥Bab ⇔A(ab)2+(A+C−B)(ab)+C≥0
14.10.2016 05:36
programjames1 wrote: I think that you can suppose x>y>z. If you do this, depending on if it's true or not, you can just change A, B, and C to fit the inequality. Also, It would be nice if you could elaborate on how ⇔Aa(a+b)+C(a+b)b≥Bab ⇔A(ab)2+(A+C−B)(ab)+C≥0 But if we don't suppose x>y>z,from this inequality we can get A^2+B^2+C^2-2AB-2AC-2BC<=0,and A,B,C >=0, it's much stronger than what Singular get.