Let a,b be real numbers. Prove the inequality 2(a4+a2b2+b4)≥3(a3b+ab3).
Problem
Source: Lithuania NMO,2010
Tags: inequalities, inequalities proposed
11.03.2012 07:49
b=0 is trivial, b≠0 devide both sides b4 and denote t=ab and above ineq turns out to be 2(t4+t2+1)≥3(t3+t)<=>(t−1)2(2t2+t+2)≥0
11.03.2012 15:01
This can be easily written as (a−b)4+a4+b4+a3b+ab3≥4a2b2 which follows by AM-GM.
11.03.2012 17:37
Different: Let a2+b2=m and ab=n; then we must show 2(m2−n2)≥3mn⟺(2m+n)(m−2n)≥0 Note that 2m+n=2a2+ab+2b2>0, and m−2n=a2−2ab+b2≥0 by AM-GM, so we are done.
11.03.2012 18:00
Generalization Let a,b,x,y be real numbers with 0<y≤3x. Prove the inequality x(a4+b4)+2ya2b2≥ (x+y)(a3b+ab3). Ion Bursuc
11.03.2012 21:38
⟺x(a−b)4+ab(3x−y)(a−b)2≥0
11.03.2012 21:51
If ab<0?
11.03.2012 23:33
Generalization 2 Let a,b,x,y be real numbers with 0<(n−1)2y≤(2n−1)x. Prove the inequality x(a2n+b2n)+2yanbn≥ (x+y)(a2n−1b+ab2n−1). Ion Bursuc
12.03.2012 02:03
Please see this: f=18((4a+b)2+15b2)(a−b)2=2a4+2a2b2+2b4−3a3b−3ab3≥0. over real numbers. and see this( use agl2010_lsos program found out): f=(a−b)2(a+b)2+12(b−a)2(a−b)2+12b2(a−b)2+12a2(a−b)2, f=(a−b)2(a+b)2+12a2(a−b)2+12b2(a−b)2+12(a−b)4, f=(a−b)2(a+b)2+14(a−b)4+12b2(a−b)2+12a2(a−b)2+14(b−a)2(a−b)2, f=54(a−b)2(a+b)2+34(b−a)2(a−b)2, f=54(a−b)2(a+b)2+14(a−b)4+12(b−a)2(a−b)2, f=54(a−b)2(a+b)2+14(b−a)2(a−b)2+12(a−b)4. BQ
12.03.2012 08:37
for ab<0, the inequality is trivial
12.03.2012 10:08
Refinement and generalization Let a,b,x,y be real numbers with 0<y≤3x. Prove the inequality x(a4+b4)+2ya2b2≥(x+y)(a3b+ab3)+3x−y4(a2−b2)2. Ion Bursuc
12.03.2012 13:32
littletush wrote: Let a,b be real numbers. Prove the inequality 2(a4+a2b2+b4)≥3(a3b+ab3). We have to prove that 4(a4+a2b2+b4)≥6(a3b+b3a). a4+a4+a2b2+a2b2≥4a3b, (AM−GM) b4+b4+a2b2+a2b2≥4ab3, (AM−GM) 2(a4+b4)≥2(a3b+ab3) (Rearrangements)
15.03.2012 03:52
a4+a2b2≥2a3b, (AM−GM) (1) b4+a2b2≥2b3a, (AM−GM) (2) (a4+b4)≥(a3b+ab3) ) (3) (1)+(2)+(3)
11.10.2013 20:36
a^4-(a^2)(b^2)+b^4=(a^2-ab+b^2)(a^2+ab+b^2). Now, a^2+ab+b^2>=3ab and a^2-ab+b^2>=(a^2+b^2)/2 (in fact simplifying and calculating gives (a-b)^2>=0) Multiply the two equations....
13.05.2017 15:47
LHS-RHS=(a−b)2(2(a2+b2)+ab)≥0
13.05.2017 17:21
littletush wrote: Let a,b be real numbers. Prove the inequality 2(a4+a2b2+b4)≥3(a3b+ab3). 2(a4+a2b2+b4)−3(a3b+ab3)=14(a−b)2{5(a+b)2+3(a−b)2}≥0.
14.05.2017 07:21
littletush wrote: Let a,b be real numbers. Prove the inequality 2(a4+a2b2+b4)≥3(a3b+ab3). Easily proved by substituting for either a+b,ab or a+b,a−b.
22.10.2017 23:45
littletush wrote: Let a,b be real numbers. Prove the inequality 2(a4+a2b2+b4)≥3(a3b+ab3). 2(a4+a2b2+b4)−3(a3b+ab3)=(a−b)2(2a2+ab+2b2)=(a−b)2{2(a+b4)2+158b2}≥0
24.10.2017 19:31
OK, so we have a^4 + b^4 >= a^3b + b^3a (really trivial, just put everything on left and you will easily prove this as you will get (a-b)(a^3-b^3) = (a-b)^2*(a^2 + ab + b^2) > = 0). Also, we have a^2 + b^2 >= 2ab, (a^2 + b^2)^2 >= 2ab(a^2 + b^2) = 2a^3b + 2b^3a Combining these two inequalities we get the solution. Man, this was easy! I don't know why is this is in High School Olympiads, but is definitely far from being a challenge, even for a beginner like me
13.11.2017 15:10
We may assume that a≥0,b≥0. By AM-GM, a4+a2b2≥2a3b, b4+a2b2≥2ab3 and 34a4+14b4≥a3b, 14a4+34b4≥ab3 Summing-up theses inequalities, we're done