Okay, write $\frac {1+abc}{\sum a}=\sum ab$. This is obvious now, $abc\leq \frac {1}{8}$ $\textstyle{\sum a}\displaystyle\geq \frac {3}{2}$

Problem. Let a, b, c be three nonnegative reals such that (b + c) (c + a) (a + b) = 1. Prove the inequality $bc+ca+ab\leq\frac34$. Solution. Assume, for the sake of contradiction, that $bc+ca+ab>\frac34$. Then, we have $bc+ca+ab=\frac34k$ for some real k > 1. Hence, if we define $A=\frac{a}{\sqrt{k}}$, $B=\frac{b}{\sqrt{k}}$ and $C=\frac{c}{\sqrt{k}}$, then we have $BC+CA+AB=\frac{b}{\sqrt{k}}\cdot\frac{c}{\sqrt{k}}+\frac{c}{\sqrt{k}}\cdot\frac{a}{\sqrt{k}}+\frac{a}{\sqrt{k}}\cdot\frac{b}{\sqrt{k}}=\frac{bc}{k}+\frac{ca}{k}+\frac{ab}{k}$ $=\frac{bc+ca+ab}{k}=\frac{\frac34k}{k}=\frac34$. On the other hand, since k > 1, we have $\sqrt{k}>1$, and thus $A=\frac{a}{\sqrt{k}}<a$, so that a > A. Similarly, b > B and c > C. The AM-GM inequality yields $\frac{BC+CA+AB}{3}\geq \sqrt[3]{BC\cdot CA\cdot AB}$. Since $BC+CA+AB=\frac34$, this rewrites as $\frac{\left(\frac34\right)}{3}\geq \sqrt[3]{BC\cdot CA\cdot AB}$; equivalently, $\frac14\geq \sqrt[3]{A^2B^2C^2}$. Taking the square root on both sides of this inequality, we get $\frac12\geq\sqrt[3]{ABC}$; after cubing, this becomes $\frac18\geq ABC$. Thus, $ABC\leq\frac18$. Also, $\left(A+B+C\right)^2-3\left(BC+CA+AB\right)=\frac12\left(\left(B-C\right)^2+\left(C-A\right)^2+\left(A-B\right)^2\right)$ $\geq 0$; thus, $\left(A+B+C\right)^2\geq 3\left(BC+CA+AB\right)$. Since $BC+CA+AB=\frac34$, this becomes $\left(A+B+C\right)^2\geq 3\cdot\frac34=\frac94$. Taking the square root of this inequality, we get $A+B+C\geq\frac32$. This all, together with a > A, b > B and c > C, yields $\left(b+c\right)\left(c+a\right)\left(a+b\right)>\left(B+C\right)\left(C+A\right)\left(A+B\right)$ $=\left(BC+CA+AB\right)\cdot\left(A+B+C\right)-ABC$ $\geq\frac34\cdot\frac32-\frac18=1$. But we were given that (b + c) (c + a) (a + b) = 1. Thus, we obtain a contradiction. Hence, our assumption $bc+ca+ab>\frac34$ was wrong, and we must have $bc+ca+ab\leq\frac34$. Problem solved. Very nice problem, by the way. Are all problems from the JBMO already posted? darij

$$1 = (a + b)(b + c)(c + a) \geqslant \frac{8} {9}(a + b + c)(ab + bc + ca) \geqslant \frac{8} {9}(ab + bc + ca)\sqrt {3\left( {ab + bc + ca} \right)}$$ So, $$\left( {ab + bc + ca} \right)^{\frac{3} {2}} \leqslant \frac{{3\sqrt 3 }} {8} \Rightarrow ab + bc + ca \leqslant \frac{3} {4}$$

This one is my proposal. By the way very nice solution,Darij.

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

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 ² :le: 2(:Sigma: x ² )+3. Now 4s ² =(x+y+z) ² so the inequality is written as 2 :Sigma: xy <= (:Sigma: x ² )+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 ² ,j ² ,k ² gives (:Sigma: i^6)+3i :^2:j :^2: k :^2: >= 2 :Sigma: i :^4: (j :^2: +k ² ). 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.

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.

blahblahblah, it's perfectly alright

siuhochung wrote: blahblahblah, it's perfectly alright I guess things get lost when Darij posts under you

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

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.

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

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 $8abc \le (a+b)(b+c)(c+a)$

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.

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. _____________________________________________________________

$1=(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc$ $\geq \sqrt{3}(ab+bc+ca)^{\frac{3}{2}}-\frac{1}{3\sqrt{3}}(ab+bc+ca)^{\frac{3}{2}}$ $=\frac{8}{3\sqrt{3}}(ab+bc+ca)^{\frac{3}{2}}$ Hence $ab+bc+ca \leq \frac{3}{4}$.

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...

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 $ab+bc+ca \geq 3(abc)^\frac{2}{3}$.

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 .

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

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 $$ab+ac+ad+bc+bd+cd\leq\frac{2}{3}$$ 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 that $$ab+ac+ad+bc+bd+cd\leq\frac{32}{27(a+b+c+d)^2}$$

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 $(a+b+c)(b+c+d)(c+d+a)(d+a+b)=1$. $$abc+bcd+cda+dab \leq\frac{16}{81(a+b+c+d)}$$

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 $$abc+bcd+cda+dab-abcd \leq\frac{44}{243(a+b+c+d)}$$

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?

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)

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

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

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

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

A direct method for my latter proposals is EV. Another one, more direct. But one must see some transformations

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 that $$ab+ac+ad+bc+bd+cd\leq\frac{32}{27(a+b+c+d)^2}$$ Method 1: Let $$b+c+d=x;c+d+a=y\ldots$$ Homogenize and get a trivial inequality, even stronger than the quoted one

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

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

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.