Find all real numbers a and b such that 2(a2+1)(b2+1)=(a+1)(b+1)(ab+1). Valentin Vornicu
Problem
Source: Romanian JBMO TST 2006, Day 3, Problem 3
Tags: trigonometry, inequalities, algebra, High school olympiad
16.05.2006 23:43
(a2+1)(1+1)≥(a+1)2 (b2+1)(1+1)≥(b+1)2 (a2+1)(b2+1)≥(ab+1)2 Multiplying these we get that 4(a2+1)2(b2+1)2≥((a+1)(b+1)(ab+1))2 but the relation fromt he hypotese tells us that these are equal. Hence we must have equality in all those 3 inegalities. We get a=b=1.
17.05.2006 06:03
Another solution : we set a=tanx,b=tany and we have 2=(sinx+cosx)(siny+cosy)(sinxsiny+cosxcosy)⇒2=2sin(x+π4)sin(y+π4)cos(y−x) From this equation if y,x∈[0,2π) we get y=x=π4 or x=π4,y=π+π4 . Both of them give a=b=1 .
17.05.2006 09:45
another method: 2(a2+1)(b2+1)=(a+1)(b+1)(ab+1) ⟺(b2−b+2)a2−a(b+1)2+(2b2−b+1)=0 Since b2−b+2=(b−12)2+74>0 So the discriminant (b+1)4−4(b2−b+2)(2b2−b+1)=−(b−1)2(7b2−2b+7)≤0 . Hence there are only real roots when b=1 and this gives us a=1
17.05.2006 10:37
shyong wrote: another method: 2(a2+1)(b2+1)=(a+1)(b+1)(ab+1) ⟺(b2−b+2)a2−a(b+1)2+(2b2−b+1)=0 Since b2−b+2=(b−12)2+74>0 So the discriminant (b+1)4−4(b2−b+2)(2b2−b+1)=−(b−1)2(7b2−2b+7)≤0 . Hence there are only real roots when b=1 and this gives us a=1
17.05.2006 11:06
kunny wrote: shyong wrote: another method: 2(a2+1)(b2+1)=(a+1)(b+1)(ab+1) ⟺(b2−b+2)a2−a(b+1)2+(2b2−b+1)=0 Since b2−b+2=(b−12)2+74>0 So the discriminant (b+1)4−4(b2−b+2)(2b2−b+1)=−(b−1)2(7b2−2b+7)≤0 . Hence there are only real roots when b=1 and this gives us a=1 anything wrong with that factorization ?? It seems ok to me
17.05.2006 12:01
Shyong, I think Kunny was puzzled (amased) how u factored this. So am I
17.05.2006 12:11
delta wrote: Shyong, I think Kunny was puzzled (amased) how u factored this. So am I actually , you can either expand everything out and do the rational roots theorem things . But somehow , from above post , i already know that b=1 is the answer . So i guess the discriminant should looks something like (b−1)2(pb2+qb+r) . Then you use substitute for some value for left and right and solve simultaneous equation to get p,q,r .
26.07.2014 17:03
(from C.B.S) (a*a+1)(1+1)>=(a+1)(a+1) (1) equality for a=1 (b*b+1)(1+1)>=(b+1)(b+1) (2) equality for b=1 (a*a+1)(b*b+1)>=(a+b)(a+b) (3) equality for a=b=1 (1)*(2)*(3)→(a*a+1)2*(b*b+1)2 *2*2>=(a+1)2*(b+1)2(a+b)2|√→ (a*a+1)(b*b+1)*2>=(a+1)(b+1)(a+b) equality for a=b=1 In conclusion the solutions of problems are a=b=1
04.06.2020 02:36
Let a,b,c be reals. Prove that(1+a2)(1+b2)(1+c2)≥(1+ab)(1+bc)(1+ca)⟹Let a,b be real numbers. Prove that2(a2+1)(b2+1)≥(a+1)(b+1)(ab+1)