Find the smallest constant k such that x√x+y+y√y+z+z√z+x≤k√x+y+z for all positive x, y, z.
Problem
Source: Jack Garfunkel, P1490, CRUX 1989/9, p.270, Chinese TST 2007 5th quiz P3
Tags: inequalities, 3-variable inequality, cyclic inequality, square root inequality
08.11.2003 17:45
The correct answer is k=54.
13.11.2003 23:16
This inequality is beautiful and very difficult to solve. I woul like to know if you Fedor created it or you found it on a mathematical review or it is an olympiad problem. Thank you.
14.11.2003 12:56
I agree that the inequality is extremely difficult and beautiful. It is an inequality that you remember after you had solved it. I saw it in Crux and I think (but I'm not sure) that it was given also at an american contest.
15.11.2003 13:43
Unfortunately, yesterday was a very bad day for me. I will post a detailed solution of this problem and I hope today I'm not so stupid. As in the above "solution", we have to prove that if a^2,b^2,c^2 are the sides of o triangle( even degenerate), then: a+b+c+(a+b+c)(a-b)(a-c)(c-b)/abc<=5/(2sqrt2)*sqrt(a^2+b^2+c^2). Proposition: The above inequality is veryfied when b<=c<=a<=sqrt(b^2+c^2). Proof: It is easy to prove that 5/(2sqrt2)*sqrt(a^2+b^2+c^2)>=5/4*(a+sqrt(b^2+c^2)). Thus, we have to prove that (a+b+c)+(a+b+c)(a-b)(a-c)(c-b)/abc<=5/4(a+sqrt(b^2+c^2)), which reduces (after clearing denominators) to proving that f(a)<=0, where f(x)=4x^3(c-b)-x^2bc+x(4b^3+4b^2c+4bc^2-4c^3-5bcsqrt(b^2+c^2))+4bc(c^2-b^2). We know that b<=c<=a<=sqrt(b^2+c^2). If b=c, proving that f(a)<=0 reduces to ab^2(a-b+(5sqrt2-7)b)>0, which is obvious because a>=b. So, we can suppose that b<c. It is clear that when x goes to minus infinity , f(x) goes to minus infinity and when x goes to infinity, f(x) goes to infinity. Also, f(0)>=0 because c>=b. It is trivial to prove that f(c)<=0 (it reduces to 5sqrt(b^2+c^2)>=4b+3c, which is Cauchy and that f(sqrt(b^2+c^2))<=0 (which reduces to bc(sqrt(b^2+c^2)-2b)^2>=0. So f has a negative (possible 0) root, a root between 0 and c and another one which is at least sqrt(b^2+c^2). Consequently, f doesn't change sign in the interval [c, sqrt(b^2+c^2)]. Since it isn't positive in c and sqrt(b^2+c^2) and c<=a<=sqrt(b^2+c^2), we must have f(a)<=0. Now let's solve the problem. One can easily see that we can suppose that c>=b (if not, we interchange b and c and the LHS will become smaller, while the RHS will remain the same). Now, if a<=b<=c, then we apply the proposition for the triplet (c,a,b) (this means that we have sqrt(b^2+a^2)>=c>=b>=a and we can appply the proposition) and if b<=a<=c ,then we apply the proposition for the triplet (c,b,a).
16.11.2003 15:48
Congratulations, Harazi, your solution is now correct. I verified it in detail. Is very nice solution for this hard problem. I also came to idea to change from (x,y,z) to (a,b,c), but can't go further. The case (a>=c>=b) is really main case in this inequality. The idea f(c) <= 0, f(sqrt(b^2+c^2)) <=0 is nice. Namdung
22.05.2006 19:11
hitek wrote: Let x,y,z be real non-negative numbers, prove that x√x+y+y√y+z+z√z+x≤54√x+y+z. This was suggested by Walther Janous's problem Crux 1366 [1989:271]. There can be someone it knows? What there was, in Crax 1366? Thank you.
23.05.2006 15:41
a√a+b+b√b+c+c√c+a≤54√a+b+c Here's my solution Notice that if one of three number a,b,c equal 0 then we are done. Otherwise, let x=√a+b2,y=√a+c2,z=√b+c2,k=5√24 Rewrite the problem into the form (x+y+z)+(x−y)(x−z)(z−y)(1xy+1yz+1zx)≤k√x2+y2+z2(∗) Clearly, we only need to solve problem incase x≥y≥z Notice that for all t≥0 then k√(x+t)2+(y+t)2+(z+t)2≥k√x2+y2+z2+3t From this result, we deduce that if we decrease all x,y,z by a number t≤min then the expression (x-y)(x-z)(z-y)(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}) is increased but the expression k\sqrt{x^2+y^2+z^2}-(x+y+z) is decreased. So after choosing a number t for which x-t,y-t,z-t is three lengths of a right triangle, we only need to consider the first problem incase abc=0, which was proved as above.
11.06.2006 20:45
This is Crux Mathematicorum, volume 17,#1, January 1991, problem 1490. Proposed by J.Garfunkel. Search in the net. There you will find the solution! Or see here post #9: http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Crux&t=1459 but it is very hard to read.
19.03.2007 07:32
Fedor Bakharev wrote: Find the smallest constant k such that \frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}\leq k\sqrt{x+y+z} for all positive x,y,z. This was proposed by Jack Garfunkel, see: Problem 1490^{*}, CRUX Mathematicorum, 1989, No.9, p.270. \sum_{cyc}\frac{x}{\sqrt{(x+y)(x+y+z)}} \leq\sum_{cyc}\frac{425x^{2}+9y^{2}+81z^{2}-54yz+810zx+514xy}{4[103(x^{2}+y^{2}+z^{2})+254(yz+zx+xy)]}=\frac{5}{4}, with equality if x=3,y=1,z=0.
19.03.2007 08:56
Any proof to go with it? Is it just another computer proof... that's a really nice identity/inequality if you can prove it. What is the equality case for \frac{425x^{2}+9y^{2}+81z^{2}-54yz+8101zx+514xy}{4[103(x^{2}+y^{2}+z^{2})+254(yz+zx+xy)]}\ge\frac{x}{\sqrt{(x+y)(x+y+z)}} Is equality at x=3y,z\to 0? edit. oops didn't see your equality case... but it is in fact for any x=3y, right?
13.06.2007 11:00
And now, here is my solution for it LHS^{2}\le \sum a(5a+b+9c)\cdot \sum \frac{a}{(a+b)(5a+b+9c)}=5(a+b+c)^{2}\cdot\sum\frac{a}{(a+b)(5a+b+9c)} Hence, it suffices to prove that (a+b+c)\cdot\sum\frac{a}{(a+b)(5a+b+9c)}\le \frac{5}{16} But we have RHS-LHS=\frac{\sum ab(a+b)(a+9b)(a-3b)^{2}+243\sum a^{3}b^{2}c+835\sum a^{3}bc^{2}+232\sum a^{4}bc+1230a^{2}b^{2}c^{2}}{16\prod (a+b) \cdot \prod (5a+b+9c)}\ge 0 We are done. @: I'm looking for you idea.
29.07.2011 19:07
hungkhtn wrote: k\sqrt{(x+t)^2+(y+t)^2+(z+t)^2} \ge k\sqrt{x^2+y^2+z^2}+3t From this result, we deduce that if we decrease all x,y,z by a number t \le \min(x,y,z) then the expression (x-y)(x-z)(z-y)(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}) is increased but the expression k\sqrt{x^2+y^2+z^2}-(x+y+z) is decreased. I am sorry,I think your solution is not very correct.Because k\sqrt{x^2+y^2+z^2}-(x+y+z) may be not decreased when x,y,z reduce be a number t \le \min(x,y,z) .
30.07.2011 02:16
Dear Zhaobin, i think the idea by mv method is correct,though it is not detailed.
13.10.2013 08:37
Fedor Bakharev wrote: Find the smallest constant k such that \frac {x}{\sqrt {x + y}} + \frac {y}{\sqrt {y + z}} + \frac {z}{\sqrt {z + x}}\leq k\sqrt {x + y + z} for all positive x, y, z. For x=3, y=1 and z\rightarrow0^+ we get k\geq\frac{5}{4}. We'll prove that: \frac {x}{\sqrt {x + y}} + \frac {y}{\sqrt {y + z}} + \frac {z}{\sqrt {z + x}}\leq \frac{5}{4}\sqrt {x + y + z} for all non-negatives x, y and z such that xy+xz+yz\neq0. By Cauchy-Schwarz \left(\sum_{cyc}\frac {x}{\sqrt {x + y}}\right)^2\leq\sum_{cyc}\frac{x(2x+4y+z)}{x+y}\sum_{cyc}\frac{x}{2x+4y+z}. Id est, it remains to prove that \sum_{cyc}\frac{x(2x+4y+z)}{x+y}\sum_{cyc}\frac{x}{2x+4y+z}\leq\frac{25(x+y+z)}{16}, which is \sum_{cyc}(8x^6y+72x^6z-14x^5y^2+312x^5z^2-92x^4y^3+74x^4z^3+ +122x^5yz+217x^4y^2z+143x^4z^2y+564x^3y^3z+1338x^3y^2z^2)\geq0, which is \sum_{cyc}2xy(4x+y)(x-3y)^2(x+2y)^2+ +\sum_{cyc}(122x^5yz+217x^4y^2z+143x^4z^2y+564x^3y^3z+1338x^3y^2z^2)\geq0, which is obvious.
10.02.2015 14:33
Can we let a=sqrt(x+y),b=sqrt(y+z),c=sqrt(z+x),and then use 2c cosB to replace (a^2-b^2+c^2)/a...and so on(it must form a triangle because x,y,z are positive).But I don't know how to do it then.
30.12.2015 19:42
hitek wrote: Let x,y,z be real non-negative numbers, prove that \frac x{\sqrt{x+y}}+\frac y{\sqrt{y+z}}+\frac z{\sqrt{z+x}}\leq\frac54\sqrt{x+y+z}. I think I've got a new solution: Let x+y=c^2,y+z=a^2,z+x=b^2, with a,b,c>0.Wlog, we suppose a\ge b,c.Then we have : x=\frac{c^2+b^2-a^2}{2},y=\frac{a^2-b^2+c^2}{2}, z=\frac{a^2+b^2-c^2}{2}. Then, we just have to prove that : \frac{c^2+b^2-a^2}{2}+\frac{a^2-b^2+c^2}{2}+\frac{a^2+b^2-c^2}{2}\le\frac{5}{2\sqrt{2}}\sqrt{a^2+b^2+c^2}, i.d (a+b+c)+\frac{(a+b+c)(a-b)(a-c)(c-a)}{abc}\le\frac{5}{2\sqrt{2}}\sqrt{a^2+b^2+c^2}. But \sqrt{a^2+b^2+c^2}\ge\frac{a+\sqrt{b^2+c^2}}{\sqrt{2}} ;So we just have to prove that f(a)\le 0, where : f(a)=4a^3(c-b)-a^2bc+a(4b^3+4b^2c-4c^3-5bc\sqrt{b^2+c^2})+4bc(c^2-b^2), where a is between max(a,b) and \sqrt{a^2+b^2}. Then we just have to prove the inequality when b\le c, i.d when b=c and b<c. -When b=c, we have : f(a)=-ab^2((a-b)+(5\sqrt{5}-7)b)<0. - When b<c, we have \lim\limits_{x \rightarrow +\inifity} f(x)=+\infinity, \lim\limits_{x \rightarrow -\infinity} f(x)=-\infinity, f(0)>0, and f(c)=-bc^2(5\sqrt{b^2+c^2}-4b-3c)\le 0. We also have : f(\sqrt{b^2+c^2})=2bc(4b\sqrt{b^2+c^2}-5b^2-c^2)=-2bc(\sqrt{b^2+c^2}-2b)^2\le 0. Thus, f admits three real roots (a negative root, a root in [0,c] and a root \ge\sqrt{b^2+c^2}. Then, f is negative in [c,\sqrt{b^2+c^2}]. We have f(a)=0 if \sqrt{b^2+c^2}=2b and a=\sqrt{b^2+c^2}, i.e the vector (a,b,c) is proportional to (2,1,\sqrt{3}). In this case, the inequality \frac{a+\sqrt{b^2+c^2}}{\sqrt{2}}\le\sqrt{a^2+b^2+c^2} is an equality. In conclusion, we have an equality iff vectors (x,y,z) and (0,3,1) are proportional.
22.12.2018 17:53
Fedor Bakharev wrote: Find the smallest constant k such that \frac {x}{\sqrt {x + y}} + \frac {y}{\sqrt {y + z}} + \frac {z}{\sqrt {z + x}}\leq k\sqrt {x + y + z} for all positive x, y, z.
13.02.2019 03:50
hitek wrote: Let x,y,z be real non-negative numbers, prove that \frac x{\sqrt{x+y}}+\frac y{\sqrt{y+z}}+\frac z{\sqrt{z+x}}\leq\frac54\sqrt{x+y+z}. Let a,b,c be real non-negative numbers, prove that \frac{\sqrt{2}}{2}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\leq\frac a{\sqrt{a+b}}+\frac b{\sqrt{b+c}}+\frac c{\sqrt{c+a}}\leq \frac54\sqrt{a+b+c}. here Walther Janous Jack Garfunkel
14.03.2019 06:05
Just try to graph this problem. WLOG let x+y+z =1 and all variables are positive (because we are looking for max). Use spherical coordinate to parameterize x,y and z: z=\sin ^2(\theta ); y=(\cos (\theta ) \cos (\phi ))^2; x=(\cos (\theta ) \sin (\phi ))^2 \text{Plot3D}\left[\frac{x}{\sqrt{x+y}}+\frac{z}{\sqrt{x+z}}+\frac{y}{\sqrt{y+z}},\left\{\phi ,-\frac{\pi }{2},\frac{\pi }{2}\right\},\{\theta ,0,2 \pi \},\text{PlotPoints}\to 50\right] Find one maximum point by: \text{FindMaximum}\left[\frac{x}{\sqrt{x+y}}+\frac{z}{\sqrt{x+z}}+\frac{y}{\sqrt{y+z}},\{\{\phi ,0\},\{\theta ,2 \pi \}\}\right] \Rightarrow 1.25
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24.09.2021 09:28
Let x,y,z> 0. Prove that \frac{x}{\sqrt{x+y}}+\frac{2y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}} \leq \frac{7}{8}\sqrt{x+y+z} \frac{x}{\sqrt{x+y}}+\frac{2y}{\sqrt{y+z}}+\frac{2z}{\sqrt{z+x}} \leq \frac{5}{2}\sqrt{x+y+z} \frac{x}{\sqrt{x+y}}+\frac{4y}{\sqrt{y+z}}+\frac{4z}{\sqrt{z+x}} \leq 5\sqrt{x+y+z} \frac{x}{\sqrt{x+y}}+\frac{y^2}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}} \leq \frac{5}{4}\sqrt{x+y+z}
19.12.2022 15:43
sqing wrote: Let x,y,z> 0. Prove that \frac{x}{\sqrt{x+y}}+\frac{y^2}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}} \leq \frac{5}{4}\sqrt{x+y+z} It's wrong. Try x=z=1 and y=100.
19.12.2022 20:18
sqing wrote: Let x,y,z> 0. Prove that \frac{x}{\sqrt{x+y}}+\frac{2y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}} \leq \frac{7}{8}\sqrt{x+y+z} It's wrong for y\rightarrow+\infty.