The correct answer is $k=\frac54$.

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.

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.

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).

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

hitek wrote: Let $x,y,z$ be real non-negative numbers, prove that $$\frac x{\sqrt{x+y}}+\frac y{\sqrt{y+z}}+\frac z{\sqrt{z+x}}\leq\frac54\sqrt{x+y+z}.$$ 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.

$$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\le \frac{5}{4}\sqrt{a+b+c}$$ Here's my solution Notice that if one of three number $a,b,c$ equal $0$ then we are done. Otherwise, let $$x=\sqrt{\frac{a+b}{2}}, y=\sqrt{\frac{a+c}{2}}, z=\sqrt{\frac{b+c}{2}},k=\frac{5\sqrt{2}}{4}$$ Rewrite the problem into the form $$(x+y+z)+(x-y)(x-z)(z-y)(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}) \le k\sqrt{x^2+y^2+z^2} (*)$$ Clearly, we only need to solve problem incase $x \ge y \ge z$ Notice that for all $t \ge 0$ then $$k\sqrt{(x+t)^2+(y+t)^2+(z+t)^2} \ge k\sqrt{x^2+y^2+z^2}+3t$$ 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. 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.

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.

I think this would help

Attachments:
hard ineq.doc (46kb)

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 $x=3,y=1,z=0$.

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?

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.

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 $t \le \min(x,y,z)$ .

Dear Zhaobin, i think the idea by mv method is correct,though it is not detailed.

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.

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.

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.

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 $x$, $y$, $z$.

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

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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Let $x,y,z> 0.$ Prove that $$\frac{x}{\sqrt{x+y}}+\frac{2y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}} \leq \frac{7}{8}\sqrt{x+y+z}$$ $$\frac{x}{\sqrt{x+y}}+\frac{2y}{\sqrt{y+z}}+\frac{2z}{\sqrt{z+x}} \leq \frac{5}{2}\sqrt{x+y+z}$$ $$\frac{x}{\sqrt{x+y}}+\frac{4y}{\sqrt{y+z}}+\frac{4z}{\sqrt{z+x}} \leq 5\sqrt{x+y+z}$$ $$\frac{x}{\sqrt{x+y}}+\frac{y^2}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}} \leq \frac{5}{4}\sqrt{x+y+z}$$

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 $y=100.$

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 $y\rightarrow+\infty.$