Let a, b, c be three positive reals such that (a+b)(b+c)(c+a)=1. Prove that the following inequality holds: ab+bc+ca≤34. Cezar Lupu
Problem
Source: Romanian JBMO TST, Day 3, Problem 8
Tags: inequalities, inequalities proposed, highschoolmath
15.05.2005 13:13
Okay, write 1+abc∑a=∑ab. This is obvious now, abc≤18 ∑a≥32
15.05.2005 13:49
Problem. Let a, b, c be three nonnegative reals such that (b + c) (c + a) (a + b) = 1. Prove the inequality bc+ca+ab≤34. Solution. Assume, for the sake of contradiction, that bc+ca+ab>34. Then, we have bc+ca+ab=34k for some real k > 1. Hence, if we define A=a√k, B=b√k and C=c√k, then we have BC+CA+AB=b√k⋅c√k+c√k⋅a√k+a√k⋅b√k=bck+cak+abk =bc+ca+abk=34kk=34. On the other hand, since k > 1, we have √k>1, and thus A=a√k<a, so that a > A. Similarly, b > B and c > C. The AM-GM inequality yields BC+CA+AB3≥3√BC⋅CA⋅AB. Since BC+CA+AB=34, this rewrites as (34)3≥3√BC⋅CA⋅AB; equivalently, 14≥3√A2B2C2. Taking the square root on both sides of this inequality, we get 12≥3√ABC; after cubing, this becomes 18≥ABC. Thus, ABC≤18. Also, (A+B+C)2−3(BC+CA+AB)=12((B−C)2+(C−A)2+(A−B)2) ≥0; thus, (A+B+C)2≥3(BC+CA+AB). Since BC+CA+AB=34, this becomes (A+B+C)2≥3⋅34=94. Taking the square root of this inequality, we get A+B+C≥32. This all, together with a > A, b > B and c > C, yields (b+c)(c+a)(a+b)>(B+C)(C+A)(A+B) =(BC+CA+AB)⋅(A+B+C)−ABC ≥34⋅32−18=1. But we were given that (b + c) (c + a) (a + b) = 1. Thus, we obtain a contradiction. Hence, our assumption bc+ca+ab>34 was wrong, and we must have bc+ca+ab≤34. Problem solved. Very nice problem, by the way. Are all problems from the JBMO already posted? darij
15.05.2005 13:59
1=(a+b)(b+c)(c+a)⩾ So, \left( {ab + bc + ca} \right)^{\frac{3} {2}} \leqslant \frac{{3\sqrt 3 }} {8} \Rightarrow ab + bc + ca \leqslant \frac{3} {4}
15.05.2005 14:17
This one is my proposal. By the way very nice solution,Darij.
15.05.2005 17:20
Siuhochung wrote \displaystyle 1 = (a + b)(b + c)(c + a) \geqslant \frac{8}{9}(a + b + c)(ab + bc + ca) \geqslant \frac{8}{9}(ab + bc + ca)\sqrt {3\left( {ab + bc + ca} \right)} \displaystyle \left( {ab + bc + ca} \right)^{\frac{3}{2}} \leqslant \frac{{3\sqrt 3 }}{8} \Rightarrow ab + bc + ca \leqslant \frac{3}{4} It's also my solution
15.05.2005 21:34
set x=a+b, y=b+c, z=c+a. Then xyz=1 and x,y,z are sides of a triange. Denote s the semiperimetre of this triangle. Then a=s-y, b=s-z, c=s-x. In what follows all the sums are cyclic. The inequality can be written as 2 :Sigma: ab :le: 3/2 or :Sigma: a(b+c) :le: 3/2 or :Sigma: x(s-x) :le: 3/2 or 4s <sup>2</sup> :le: 2(:Sigma: x <sup>2</sup> )+3. Now 4s <sup>2</sup> =(x+y+z) <sup>2</sup> so the inequality is written as 2 :Sigma: xy <= (:Sigma: x <sup>2</sup> )+3. Set x=i :^3: ,y=j :^3: , z=k :^3: . Then ijk=1 and the inequality becomes (:Sigma: i^6)+3i :^2:j :^2: k :^2: :ge: 2( :Sigma: i :^3: j :^3: ) which can follow easily by Schur's inequality. Indeed for n=1 Schur's ineq for i <sup>2</sup> ,j <sup>2</sup> ,k <sup>2</sup> gives (:Sigma: i^6)+3i :^2:j :^2: k :^2: >= 2 :Sigma: i :^4: (j :^2: +k <sup>2</sup> ). The rest follows by summing inequalitities i :^4: j :^2: +j :^4: i :^2: >= 2i :^3: j :^3: , etc. Well a bit complicated but it does the trick i hope.
15.05.2005 22:09
Was my original solution incorrect? (I know it was a little brief, but still...) (x+y)(y+z)(z+x)=(x+y+z)(xy+yz+zx)-xyz. Hence \frac {1+xyz}{x+y+z}=xy+yz+zx By the condition, we see that xyz\leq \frac 18 Similarly, since (x+y+z)^3=\geq 4\sum x^2y+3xyz\geq \frac {27}{8}(x+y)(y+z)(z+x) by one application of Schur and AM-GM, we see x+y+z\geq \frac {3}{2} and the result follows.
15.05.2005 22:35
blahblahblah, it's perfectly alright
15.05.2005 23:18
siuhochung wrote: blahblahblah, it's perfectly alright I guess things get lost when Darij posts under you
16.05.2005 00:53
i am sorry guys but how did you get directly that : (a+b)(b+c)(c+a)>=8/9(a+b+c)(ab+ac+bc) i am really bad at inequalities
16.05.2005 08:56
maskman wrote: i am sorry guys but how did you get directly that : (a+b)(b+c)(c+a)>=8/9(a+b+c)(ab+ac+bc) i am really bad at inequalities Expand and amgm.
16.05.2005 13:08
blahblahblah wrote: siuhochung wrote: blahblahblah, it's perfectly alright I guess things get lost when Darij posts under you I'm really sorry for this effect; actually, it took me more than an hour in total to write down the solution, since I had my breakfast inbetween, so when I submitted it I didn't see the reply you posted half an hour before. Anyway, two solutions are better than one, I think. maskman wrote: i am sorry guys but how did you get directly that : (a+b)(b+c)(c+a)>=8/9(a+b+c)(ab+ac+bc) This is a known inequality, you can find it mentioned by Harazi as "trivial, but hyper-useful" in http://www.mathlinks.ro/Forum/viewtopic.php?t=27766 post #5 and proved in post #9. Darij
16.05.2005 16:39
maskman wrote: i am sorry guys but how did you get directly that : (a+b)(b+c)(c+a)>=8/9(a+b+c)(ab+ac+bc) i am really bad at inequalities Just write (a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc and use the well-known inequality 8abc \le (a+b)(b+c)(c+a)
16.05.2005 18:52
darij grinberg wrote: I'm really sorry for this effect; actually, it took me more than an hour in total to write down the solution, since I had my breakfast inbetween, so when I submitted it I didn't see the reply you posted half an hour before. Anyway, two solutions are better than one, I think. I was just kidding, anyways. Nice solution, by the way.
01.07.2006 21:08
Q: a,b,c are reals > 0 and (a+b)(b+c)(c+a) = 1. Show: ab + bc + ca <= 3/4. A: Let x = a+b, y = b+c , and z = c+a then we obtain the following: i) x+y > z ; y+z > x ; and z+x > y. ii) a = (x + z - y)/2 ; b = (x + y - z)/2 ; c = (y + z - x)/2. iii) xyz = 1. iv) x,y, and z > 0. Then we get the equivalent ineq: 2xy + 2yz + 2zx - x^2 - y^2 - z^2 - 3 <= 0. (*) subject to the constraints: v) x,y, and z > 0. vi) xyz = 1. vii) x+y > z, y+z > x, and z+x > y. Using vi) (*) becomes: 2/x + 2/y + 2/z - x^2 -y^2 -z^2 - 3 <= 0. (**) Let f(x,y,z) = 2/x + 2/y + 2/z - x^2 -y^2 -z^2 - 3. We show that the maximum value of f is 0 under the constraints v), vi), and vii). So write g(x,y,z) = xyz = 1. Use Lagrange Multipliers, we have: (df/dx,df/dy,df/dz) = r(dg/dx,dg/dy,dg/dz) for some r. So: -2/x^2 - 2x = ryz. (1) -2/y^2 - 2y = rzx. (2) -2/z^2 - 2z = rxy. (3) Multiply (1), (2), and (3) by x, y, and z respectively we get: -2/x - 2x^2 = rxyz = r -2/y - 2/y^2 = rxyz = r -2/z - 2z^2 = rxyz = r Thus: 1/x + x^2 = 1/y + y^2 = 1/z + z^2. ( = -r/2 ) (**) We show that system (**) under the condition vii) above have only one solution (x,y,z) = (1,1,1). The first equation of (**) gives: (y - x)(1 - xy(x+y)) = 0. So if y < x or y > x, then 1 - xy(x+y) = 0 or 1 = xy(x+y) > xyz = 1 by vii). Thus we obtain a contradiction. So x = y. Similarly the second equation in (**) yields: y = z. Together we have: x = y = z = 1 since xyz = 1. Thus the maximum value of f occurs at (1,1,1) and one can check that f(1,1,1) = 0 and this concludes the proof. _____________________________________________________________
03.07.2006 10:56
1=(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc \geq \sqrt{3}(ab+bc+ca)^{\frac{3}{2}}-\frac{1}{3\sqrt{3}}(ab+bc+ca)^{\frac{3}{2}} =\frac{8}{3\sqrt{3}}(ab+bc+ca)^{\frac{3}{2}} Hence ab+bc+ca \leq \frac{3}{4}.
03.07.2006 12:44
Sung-yoon Kim wrote: ab+bc+ca)-abc \geq \sqrt{3}(ab+bc+ca)^{\frac{3}{2}}-\frac{1}{3\sqrt{3}}(ab+bc+ca)^{\frac{3}{2}} Annyonghaseyo... I don't understand this part...
03.07.2006 12:48
yookon88 wrote: Sung-yoon Kim wrote: ab+bc+ca)-abc \geq \sqrt{3}(ab+bc+ca)^{\frac{3}{2}}-\frac{1}{3\sqrt{3}}(ab+bc+ca)^{\frac{3}{2}} Annyonghaseyo... I don't understand this part... Oops who are you? I used (a+b+c)^{2}\geq 3(ab+bc+ca) and ab+bc+ca \geq 3(abc)^\frac{2}{3}.
03.07.2006 13:36
well , just use 9(a+b)(b+c)(a+c)\geq 8(a+b+c)(ab+bc+ca) and \frac{2(a+b+c)}{3}\geq \sqrt[3]{(a+b)(b+c)(c+a)} and we are done .
11.01.2022 05:42
sqing wrote: Let a,b,c,d be positive reals such that (a+b+c)(b+c+d)(c+d+a)(d+a+b)=1. Prove that abc+bcd+cda+dab \leq\frac{4}{27} corrected. Weaker than Turkevich
11.01.2022 19:18
sqing wrote: Let a,b,c,d be positive reals such that (a+b+c)(b+c+d)(c+d+a)(d+a+b)=1. Prove thatab+ac+ad+bc+bd+cd\leq\frac{2}{3} This is stronger: Let a,b,c,d be positive reals such that (a+b+c)(b+c+d)(c+d+a)(d+a+b)=1. Prove thatab+ac+ad+bc+bd+cd\leq\frac{32}{27(a+b+c+d)^2}
11.01.2022 19:21
sqing wrote: Let a,b,c,d be positive reals such that (a+b+c)(b+c+d)(c+d+a)(d+a+b)=1. abc+bcd+cda+dab \leq\frac{4}{27} This is much stronger: Let a,b,c,d be positive reals such that (a+b+c)(b+c+d)(c+d+a)(d+a+b)=1. abc+bcd+cda+dab \leq\frac{16}{81(a+b+c+d)}
12.01.2022 05:56
sqing wrote: Let a,b,c,d be positive reals such that (a+b+c)(b+c+d)(c+d+a)(d+a+b)=1. Prove that abc+bcd+cda+dab-abcd \leq\frac{11}{81} Let a,b,c,d be positive reals such that (a+b+c)(b+c+d)(c+d+a)(d+a+b)=1. Prove that abc+bcd+cda+dab-abcd \leq\frac{44}{243(a+b+c+d)}
12.01.2022 07:04
sqing wrote: sqing wrote: Let a,b,c,d be positive reals such that (a+b+c)(b+c+d)(c+d+a)(d+a+b)=1. Prove that abc+bcd+cda+dab-abcd \leq\frac{11}{81} Let a,b,c,d be positive reals such that (a+b+c)(b+c+d)(c+d+a)(d+a+b)=1. Prove that abc+bcd+cda+dab-abcd \leq\frac{44}{243(a+b+c+d)} Are you sure?
12.01.2022 07:22
sqing wrote: sqing wrote: Let a,b,c,d be positive reals such that (a+b+c)(b+c+d)(c+d+a)(d+a+b)=1. Prove that abc+bcd+cda+dab-abcd \leq\frac{11}{81} Let a,b,c,d be positive reals such that (a+b+c)(b+c+d)(c+d+a)(d+a+b)=1. Prove that abc+bcd+cda+dab-abcd \leq\frac{44}{243(a+b+c+d)} If t\approx0.151 is the unique positive real solution to 3t(2t+1)^3=1, then a=b=c=t and d=1 is a counterexample. (found with computer)
12.01.2022 09:25
bel.jad5 wrote: TropicalDog wrote: Have (a+b)(b+c)(c+a)=1 Prove that ab+bc+ca\leq\frac{3}{4} Clearly a+b+c\geq \frac{3}{2} and abc \leq \frac{1}{8} and we have: ab+bc+ac \leq \frac{2}{3}(a+b+c)(ab+bc+ca)=\frac{2}{3}(1+abc) \leq \frac{2}{3}(1+\frac{1}{8})=\frac{3}{4} https://artofproblemsolving.com/community/c6h1325028p7136730 https://artofproblemsolving.com/community/c6h572665p3368477 https://artofproblemsolving.com/community/c6h378313p2090115 https://artofproblemsolving.com/community/c6h1184603p5748506
12.01.2022 17:57
jasperE3 wrote: sqing wrote: sqing wrote: Let a,b,c,d be positive reals such that (a+b+c)(b+c+d)(c+d+a)(d+a+b)=1. Prove that abc+bcd+cda+dab-abcd \leq\frac{11}{81} Let a,b,c,d be positive reals such that (a+b+c)(b+c+d)(c+d+a)(d+a+b)=1. Prove that abc+bcd+cda+dab-abcd \leq\frac{44}{243(a+b+c+d)} If t\approx0.151 is the unique positive real solution to 3t(2t+1)^3=1, then a=b=c=t and d=1 is a counterexample. (found with computer) Bravo
13.01.2022 15:32
mihaig wrote: This is much stronger: Let a,b,c,d be positive reals such that (a+b+c)(b+c+d)(c+d+a)(d+a+b)=1. abc+bcd+cda+dab \leq\frac{16}{81(a+b+c+d)} I have 2 methods of proving. So te problem is, of course, correct. Not incorrect
13.01.2022 21:38
mihaig wrote: Let a,b,c,d be positive reals such that (a+b+c)(b+c+d)(c+d+a)(d+a+b)=1. abc+bcd+cda+dab \leq\frac{16}{81(a+b+c+d)} Same here. There are 3 proofs. Correct and easy
16.01.2022 22:41
A direct method for my latter proposals is EV. Another one, more direct. But one must see some transformations
17.01.2022 15:24
mihaig wrote: This is stronger: Let a,b,c,d be positive reals such that (a+b+c)(b+c+d)(c+d+a)(d+a+b)=1. Prove thatab+ac+ad+bc+bd+cd\leq\frac{32}{27(a+b+c+d)^2} Method 1: Let b+c+d=x;c+d+a=y\ldotsHomogenize and get a trivial inequality, even stronger than the quoted one
17.01.2022 15:42
mihaig wrote: This is much stronger: Let a,b,c,d be positive reals such that (a+b+c)(b+c+d)(c+d+a)(d+a+b)=1. abc+bcd+cda+dab \leq\frac{16}{81(a+b+c+d)} It's just Mac Laurin. A trivial form
18.01.2022 10:57
Like I said, it's just \frac14\cdot\sum{abc}\leq\frac14\cdot\sum{a}\cdot\frac16\cdot\sum{ab}.But even so, it's much stronger than the starting inequality
02.05.2022 21:12
Note that (a+b+c)(ab+bc+ca) = (a+b)(b+c)(c+a) + abc = 1 + abc and (a+b)(b+c)(c+a) \le 8abc \implies abc \le \frac{1}{8} and (a+b) + (b+c) + (c+a) \ge 3\sqrt[3]{(a+b)(b+c)(c+a)} \implies a+b+c \ge \frac{3}{2} Now Note that we need to prove 3(a+b+c) \ge 4(1+abc) which now is obvious.