The left part is easy. The term is larger than $ \frac{a}{\sqrt{a^2+b^2}}+\frac{c}{\sqrt{c^2+a^2}}$. We can assume $ c\geq b$. then it is larger than $ \frac{a}{\sqrt{a^2+c^2}}+\frac{c}{\sqrt{c^2+a^2}} > \frac{a}{a+c}+\frac{c}{a+c} = 1$. Misha

How about if $b>c$?

The inequality is cyclic so we can assume that c is the largest of a,b, c. In case of b>c you only have to replace all a's by c's all b's by a's and all c's by b's in my proof. Misha

Simpler solution for the left part: $ S > \frac{a}{\sqrt{a^2+b^2+c^2}} + \frac{b}{\sqrt{a^2+b^2+c^2}} + \frac{c}{\sqrt{a^2+b^2+c^2}}$.

Here is another proof of the left side (remember Engel ;-) ): We have $\sqrt{a^{2}+b^{2}}<\sqrt{a^{2}+b^{2}+c^{2}+2bc+2ca+2ab}=a+b+c$, and thus $\frac{a}{\sqrt{a^{2}+b^{2}}}>\frac{a}{a+b+c}$. Similarly, $\frac{b}{\sqrt{b^{2}+c^{2}}}>\frac{b}{a+b+c}$ and $\frac{c}{\sqrt{c^{2}+a^{2}}}>\frac{c}{a+b+c}$. Thus, $\frac{a}{\sqrt{a^{2}+b^{2}}}+\frac{b}{\sqrt{b^{2}+c^{2}}}+\frac{c}{\sqrt{c^{2}+a^{2}}}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1$. As for the right side, after the substitution $A=a^2$, $B=b^2$, $C=c^2$, it becomes $\frac{\sqrt{A}}{\sqrt{A+B}}+\frac{\sqrt{B}}{\sqrt{B+C}}+\frac{\sqrt{C}}{\sqrt{C+A}}\leq \frac{3\sqrt{2}}{2}$, or, equivalently, $\sqrt{\frac{A}{A+B}}+\sqrt{\frac{B}{B+C}}+\sqrt{\frac{C}{C+A}}\leq \frac{3\sqrt{2}}{2}$, what can be rewritten as $\sqrt{\frac{2A}{A+B}}+\sqrt{\frac{2B}{B+C}}+\sqrt{\frac{2C}{C+A}}\leq 3$. And this is exactly Vasc's bloody inequality 113 in Titu Andreescu, Vasile Cîrtoaje, Gabriel Dospinescu, Mircea Lascu, Old and New Inequalities, GIL, Zalau 2004. Bloody since I have been trying to solve it for an hour until I realized that it's in that book with two solutions, each one page long. Darij

I've seen quite a number of inequalities now in the form of \[ \sum_{i=1}^{n} \frac{1}{\sqrt{1+x_i}} \leq S \] where $x_1 x_2 \cdots x_n$ is a constant (often 1, as in this case). If there a general (and neat) way to tackle them?

Darij, could you please post one of the two solutions you mention because I don't have the book (or just mention which inequalities were used in the solution). Misha

I have just proved this inequality in excellent way, in my opinion. We can write \[ \sqrt\frac{a}{a+b}+\sqrt\frac{b}{b+c}+\sqrt\frac{c}{c+a}= \] \[ =\frac{a+c}{2(a+b+c)}\cdot\sqrt\frac{4a(a+b+c)^2}{(a+b)(a+c)^2}+ \frac{b+a}{2(a+b+c)}\cdot\sqrt\frac{4b(a+b+c)^2}{(b+c)(b+a)^2}+ \frac{c+b}{2(a+b+c)}\cdot\sqrt\frac{4b(a+b+c)^2}{(c+a)(c+b)^2}=S. \] Now we apply Jensen's inequality for $\sqrt{x}$ and obtain: \[ S\leq \sqrt{\frac{2a(a+b+c)}{(a+b)(a+c)}+\frac{2b(a+b+c)}{(b+c)(b+a)}+\frac{2c(a+b+c)}{(c+a)(c+b)}}=\sqrt{S'}. \] So we are to show that $S'\leq \frac{9}{2}$. You can easily check this inequality, because it is equivalent to $\sum\limits_{\textrm{cyclic}}a^2b\geq 6abc$. Is this solution distinct from Vasc's ones? P.S. Are there copyrights for solutions? Each author merited to be mentioned if his solution is used somewhere.

Myth wrote: Is this solution distinct from Vasc's ones? Not only distinct; it is also two times shorter than any of the solutions in the book, and much more elegant than any of them. Myth wrote: P.S. Are there copyrights for solutions? I hope not, but in any case, when I will refer to your solution, I will call it the Myth solution! Darij

darij grinberg wrote: Myth wrote: Is this solution distinct from Vasc's ones? Not only distinct; it is also two times shorter than any of the solutions in the book, and much more elegant than any of them. Jensen forever! darij grinberg wrote: Myth wrote: P.S. Are there copyrights for solutions? I hope not, but in any case, when I will refer to your solution, I will call it the Myth solution! I hope all people will respect solution's authors as well as you respect...

Myth wrote: darij grinberg wrote: Myth wrote: Is this solution distinct from Vasc's ones? Not only distinct; it is also two times shorter than any of the solutions in the book, and much more elegant than any of them. Jensen forever! More precisely, weighted Jensen forever! I myself had learnt it, as a method, from Peter Scholze in Athens (the inequality itself was known to me before, but I didn't really know what do with it ), and it was immediately clear to me that it should be much more powerful that plain Jensen and it should have a lot of interesting applications. But until now I have seen only very few ones, including your ingenious proof of Vasc's inequality. I guess that most applications of the weighted Jensen inequality remain undiscovered just because people too seldomly try to apply it. And don't forget the weighted Karamata inequality (I believe it's also called Fuchs inequality) - I can't believe that it has no use! Darij

A very elegant and neat solution. Congratulations, Myth.

darij grinberg wrote: Myth wrote: darij grinberg wrote: Myth wrote: Is this solution distinct from Vasc's ones? Not only distinct; it is also two times shorter than any of the solutions in the book, and much more elegant than any of them. Jensen forever! darij grinberg wrote: More precisely, weighted Jensen forever! I myself had learnt it, as a method, from Peter Scholze in Athens (the inequality itself was known to me before, but I didn't really know what do with it ), and it was immediately clear to me that it should be much more powerful that plain Jensen and it should have a lot of interesting applications. But until now I have seen only very few ones, including your ingenious proof of Vasc's inequality. I guess that most applications of the weighted Jensen inequality remain undiscovered just because people too seldomly try to apply it. And don't forget the weighted Karamata inequality (I believe it's also called Fuchs inequality) - I can't believe that it has no use! I very often use Jensen's inequality (at least, I try such approch in almost each inequalities).

What is the "weighted Karamata inequality "??

Truly beautiful and impecable! Congratulations, Myth! And it is really sincere!

anothor solution: let x=b/a, y=c/b, z=a/c, xyz=1,ln x+ln y+ln z=0 let w(t)=(1+e^(2t))^(-1/2) w"(t)>0 Jensen : (1+e^(2ln x))^(-1/2)+(1+e^(2ln y))^(-1/2)+(1+e^(2ln z))^(-1/2) <=3(1+e^(2/3*(ln x+ln y+ln z)))^(-1/2) =3*2(-0.5) (1+e^(2ln x))^(-1/2)=(1+ x^2)^(-0.5) ... (1+ x^2)^(-0.5) +(1+ y^2)^(-0.5) +(1+ z^2)^(-0.5) <=3*2(-0.5) 7615

zxingtan wrote: anothor solution: let x=b/a, y=c/b, z=a/c, xyz=1,ln x+ln y+ln z=0 let w(t)=(1+e^(2t))^(-1/2) w"(t)>0 Jensen : (1+e^(2ln x))^(-1/2)+(1+e^(2ln y))^(-1/2)+(1+e^(2ln z))^(-1/2) <=3(1+e^(2/3*(ln x+ln y+ln z)))^(-1/2) =3*2(-0.5) (1+e^(2ln x))^(-1/2)=(1+ x^2)^(-0.5) ... (1+ x^2)^(-0.5) +(1+ y^2)^(-0.5) +(1+ z^2)^(-0.5) <=3*2(-0.5) You are wrong, because your solution requires $w(t)$ is convex for all $t\in\mathbb{R}$, but it is not true.

Another nice solution is to consider the incircle of a triangle with sides $a^2+b^2$, $b^2+c^2$ and $c^2+a^2$.

I think this is more difficult. If $a,b,c$ are non-negative numbers, then $\sqrt{\frac{a}{4a+5b}}+\sqrt{\frac{b}{4b+5c}}+\sqrt{\frac{c}{4c+5a}}\leq 1$.

Ji Chen wrote: Vasc wrote: If $ a,b,c$ are positive numbers, then $ \sqrt {\frac {a}{4a + 5b}} + \sqrt {\frac {b}{4b + 5c}} + \sqrt {\frac {c}{4c + 5a}}\leq 1$.

shalex wrote: For all $ x,y,z\in\mathbb{R_{ + }}$ such that $ xyz = 1$ and $ 0\leqslant\lambda\leqslant\frac {5}{4}$ there is $ \frac {1}{\sqrt {1 + \lambda x}} + \frac {1}{\sqrt {1 + \lambda y}} + \frac {1}{\sqrt {1 + \lambda z}}\leqslant\frac {3}{\sqrt {1 + \lambda}}.$ The solution is nearly the same as Arwen's to the original one in which $ \lambda = 1$. $ \lambda > \frac {5}{4}\Longrightarrow \frac {1}{\sqrt {1 + \lambda x}} + \frac {1}{\sqrt {1 + \lambda y}} + \frac {1}{\sqrt {1 + \lambda z}} < 2.$ holds for all positive numbers $ x,y,z$ with $ xyz = 1.$

See also here : http://www.mathlinks.ro/Forum/viewtopic.php?t=106970 Let $a,b,c$ be nonnegative reals. Prove $$\sqrt{\frac a {4a+5b}}+\sqrt{\frac b {4b+5c}}+\sqrt{\frac c {4c+5a}} \ge \frac 1 2$$If $a, b, c>0.$ Prove that \[\sqrt{\frac{a}{4b+5c}}+\sqrt{\frac{b}{4c+5a}}+\sqrt{\frac{c}{4a+5b}}\ge\frac{2}{3}+\frac{ab+bc+ca}{(a+b+c)^2}\]If $a,b,c$ are non-negative numbers such that $ab+bc+ca>0.$ Find the minimum value of $\sqrt{\frac{a}{4b+9c}}+\sqrt{\frac{b}{4c+9a}}+\sqrt{\frac{c }{4a+9b}} .$ here

https://mp.weixin.qq.com/s?__biz=MzU0NDI5MjczNw==&mid=2247484665&idx=1&sn=9865704acf40177b35fa1eeceb341c51&chksm=fb7f2e17cc08a701c94da2c23cc1584bf53c67a029af8f0a3be098afb756c3ae357736c94cbd&mpshare=1&scene=1&srcid=04284qiU9DgVSHmU5DKkB7c8&pass_ticket=PddpMA1lGncTsH%2FKvyp0%2BXrxQc6%2B3VJ%2Fgv9onzl3RWY%3D#rd

Suppose that $ a$, $ b$, $ c$ are positive real numbers. Prove that $$ \frac {a}{\sqrt {a^{2} + b^{2}}} + \frac {b}{\sqrt {b^{2} + c^{2}}} + \frac {c}{\sqrt {c^{2} + a^{2}}}+\frac{\sqrt{2}}{16}\left(\frac{b^4}{a^4}+\frac{c^4}{b^4}+\frac{a^4}{c^4}\right)\geq\frac {27\sqrt {2}}{16}.$$IMO2001 2008 Jiangxi Higher Education Entrance Examination High-School Mathematics(China Tianjin) No.8(2007)：

Myth wrote: I have just proved this inequality in excellent way, in my opinion. We can write \[ \sqrt\frac{a}{a+b}+\sqrt\frac{b}{b+c}+\sqrt\frac{c}{c+a}= \]\[ =\frac{a+c}{2(a+b+c)}\cdot\sqrt\frac{4a(a+b+c)^2}{(a+b)(a+c)^2}+ \frac{b+a}{2(a+b+c)}\cdot\sqrt\frac{4b(a+b+c)^2}{(b+c)(b+a)^2}+ \frac{c+b}{2(a+b+c)}\cdot\sqrt\frac{4b(a+b+c)^2}{(c+a)(c+b)^2}=S. \]Now we apply Jensen's inequality for $\sqrt{x}$ and obtain: \[ S\leq \sqrt{\frac{2a(a+b+c)}{(a+b)(a+c)}+\frac{2b(a+b+c)}{(b+c)(b+a)}+\frac{2c(a+b+c)}{(c+a)(c+b)}}=\sqrt{S'}. \]So we are to show that $S'\leq \frac{9}{2}$. You can easily check this inequality, because it is equivalent to $\sum\limits_{\textrm{cyclic}}a^2b\geq 6abc$. Is this solution distinct from Vasc's ones? P.S. Are there copyrights for solutions? Each author merited to be mentioned if his solution is used somewhere. indeed very nice!

Suppose that $ a$, $ b$, $ c$ , $ d$ are positive real numbers, prove that \[ \frac {a}{\sqrt {a^{2} + b^{2}}} + \frac {b}{\sqrt {b^{2} + c^{2}}} + \frac {c}{\sqrt {c^{2} + d^{2}}}+ \frac {d}{\sqrt {d^{2} + a^{2}}}\leq3.\]

Vasc wrote: I think this is more difficult. If $a,b,c$ are non-negative numbers, then $\sqrt{\frac{a}{4a+5b}}+\sqrt{\frac{b}{4b+5c}}+\sqrt{\frac{c}{4c+5a}}\leq 1$. Let $\frac{a}{4a+5b}=\frac{p^2}{9}$, $\frac{b}{4b+5c}=\frac{q^2}{9}$ and $\frac{c}{4c+5a}=\frac{r^2}{9}$ Hence, we need to prove that $p+q+r\leq3$ and since $\frac{5b}{a}=(-4+\frac{9}{p^2})$, we obtain $\{p,q,r\}\in(0;\frac{3}{2})$ Let $p+q+r>3$, $p=kx$, $q=ky$ and $r=kz$, where $k>0$ and $x+y+z=3$. Hence, $k>1$, $\{x,y,z\}\in\{0;\frac{3}{2}\}$ and $125=\prod (-4+\frac{9}{k^2x^2})<\prod (-4+\frac{9}{x^2})$, which is contradiction because we'll prove now that: $125\geq \prod (-4+\frac{9}{x^2})$ Let $3u=x+y+z$, $3v^2=xy+yz+zx$ and $w^3=xyz$ Since, we need to prove $f(w^3)\geq0$, where $f(w^3)$ is increasing function. Hence, $f(w^3)$ reaches minimum value for minimum value of $w^3$, which happens when two of variables are equal. $x=y$ $27(x-1)^2(7x^2+14x+9)\geq0$

mecrazywong wrote: Suppose that $ a$, $ b$, $ c$ are positive real numbers, prove that \[ 1 < \frac {a}{\sqrt {a^{2} + b^{2}}} + \frac {b}{\sqrt {b^{2} + c^{2}}} + \frac {c}{\sqrt {c^{2} + a^{2}}}\leq\frac {3\sqrt {2}}{2} \] first is my problem

xzlbq wrote: Let $x,y,z>0$,prove \[2\,\sqrt {{\frac { \left( x+y+z \right) \left( xy+xz+yz \right) }{ \left( y+z \right) \left( z+x \right) \left( x+y \right) }}}\geq \sqrt {{\frac {x}{x+y}}}+\sqrt {{\frac {y}{y+z}}}+\sqrt {{\frac {z}{z+x}}}\]<=> \[2\,\sqrt { \left( x+y+z \right) \left( xy+xz+yz \right) }\geq \sqrt {x \left( y+z \right) \left( z+x \right) }+\sqrt {y \left( z+x \right) \left( x+y \right) }+\sqrt {z \left( x+y \right) \left( y+z \right) }\] Note this a=a: \[4\,{\frac { \left( x+y+z \right) \left( xy+xz+yz \right) }{ \left( x+ y \right) \left( y+z \right) \left( z+x \right) }}={\it \sum} \left( z+x \right) {\it \sum} \left( {\frac {x}{ \left( z+x \right) \left( x+y \right) }} \right) \] By C-S, is easy

xzlbq wrote: mecrazywong wrote: Suppose that $ a$, $ b$, $ c$ are positive real numbers, prove that \[ 1 < \frac {a}{\sqrt {a^{2} + b^{2}}} + \frac {b}{\sqrt {b^{2} + c^{2}}} + \frac {c}{\sqrt {c^{2} + a^{2}}}\leq\frac {3\sqrt {2}}{2} \] first is my problem Let $\frac{x}{x+y}=\frac{p}{2}$, $\frac{y}{y+z}=\frac{q}{2}$ and $\frac{z}{z+x}=\frac{r}{2}$ Hence, we need to prove that $p^2+q^2+r^2\geq3$ and since $\frac{y}{x}=\frac{2}{p}-1$, we obtain $\{p,q,r\}\in\{0;2\}$ Let $p^2+q^2+r^2<3$, $p=ka$, $q=kb$, $r=kc$, where $k>0$ and $a^2+b^2+c^2=3$ Hence, $k<1$ $1=\prod (\frac{2}{p}-1)=\prod (\frac{2}{ka}-1)>\prod (\frac{2}{a}-1)$,which is contradiction because we'll prove now that: $\prod (\frac{2}{a}-1)\geq 1$ or $\prod (-a+2)(-b+2)(-c+2)\geq abc$ Let $3u=a+b+c$, $3v^2=ab+bc+ca$ and $w^3=abc$, where $3u^2-2v^2=1$, then we need to prove $2w^3\leq 6v^2-12u+8=9u^2-12u+5$ Since $2w^3\leq 2(3uv^2-2u^3+2\sqrt{(u^2-v^2)^3})=5u^3-3u^2+\sqrt{2(-u^2+1)^3}$, so enough to prove $5u^3-3u^2+\sqrt{2(-u^2+1)^3}\leq 9u^2-12u+5$ or $\sqrt{2(-u^2+1)^3}\leq -5u^3+12u^3-12u+5=(-u+1)(5u^2-7u+5)$ squaring $2(-u+1)^3(u+1)^3\leq (-u+1)^2(5u^2-7u+5)^2$ or $0\leq (u-1)^2(27u^4-66u^3+99u^2-74u+23)$

Suppose that $ a$, $ b$, $ c$ are positive real numbers, prove that \[ \frac {a}{\sqrt {a^{2} + b^{2}}} + \frac {b}{\sqrt {b^{2} + c^{2}}} + \frac {c}{\sqrt {c^{2} + a^{2}}}\leq\frac {3\sqrt {2}}{2} \]$$\iff$$Suppose that $ a$, $ b$, $ c$ are positive real numbers such that $(1-a)(1-b)(1-c)=abc.$ Prove that$$\sqrt{a}+\sqrt{b}+\sqrt{c}\leq \frac{3\sqrt{2}}{2}$$here

sqing wrote: Suppose that $ a$, $ b$, $ c$ are positive real numbers, prove that $$\frac{a}{\sqrt{b^2+c^2}}+\sqrt{2}\left(\frac{b}{a+c}+\frac{c}{a+b}\right) \ge \frac{3\sqrt{2}}{2}$$ $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{\sqrt{2(a^2+b^2)}} \ge \frac{3}{2}$$Suppose that $ a$, $ b$, $ c$ are positive real numbers. Prove that $$\frac{a}{b+c}+\frac{b}{c+a}+\sqrt{\frac{2c}{a+b}} \ge {2}$$here: $$ \frac{a}{b+c}+\frac{b}{c+a}+\frac{4(c+\sqrt{ab})}{a+b}\geq 4$$ $$ \frac{a}{b+c}+\frac{b}{c+a}\ge \frac{a+b}{2}\left(\frac{1}{b+c}+\frac{1}{c+a}\right)$$

sqing wrote: High-School Mathematics(China Tianjin) No.8(2007)： If $a>0$, $b>0$, $c>0$, and $\sqrt[3]{\!abc\mspace{0.5mu}}\ge8$, then $\frac1{\sqrt{\mspace{-1mu}1+a}\mspace{1mu}}+\frac1{\sqrt{\mspace{-1mu}1+b}\mspace{1mu}}+\frac1{\sqrt{\mspace{-1mu}1+c}\mspace{1mu}}\ge\frac3{\sqrt{\mspace{-2mu}1+\sqrt[3]{\!abc\mspace{0.5mu}}\mspace{1mu}}}$?

Can we prove that suppose $x_1,\dots ,x_n>0,n\ge3$ then $\frac {x_1}{\sqrt {{x_2}^{2} + \dots +{x_n}^{2}}}+\frac {x_2}{\sqrt {{x_1}^{2} +{x_3}^{2}+ \dots +{x_n}^{2}}}+\dots +\frac {x_n}{\sqrt {{x_1}^{2} + \dots +{x_{n-1}}^{2}}} \le t$ What's the max value of t?

sqing wrote: Suppose that $ a$, $ b$, $ c$ are positive real numbers, prove that \[ \frac {a}{\sqrt {a^{2} + b^{2}}} + \frac {b}{\sqrt {b^{2} + c^{2}}} + \frac {c}{\sqrt {c^{2} + a^{2}}}\leq\frac {3\sqrt {2}}{2} \]$$\iff$$Suppose that $ a$, $ b$, $ c$ are positive real numbers such that $(1-a)(1-b)(1-c)=abc.$ Prove that$$\sqrt{a}+\sqrt{b}+\sqrt{c}\leq \frac{3\sqrt{2}}{2}$$here Our key idea is to make this geometrically, let $O$ be the origin of the complex plane. Consider points $c,-ci, -a,bi$ as $P,Q,R,S$ respectively. Note that our inequality is simply equivalent to proving that $1<\sin \angle OSR+\sin\angle OPS+\sin \angle ORQ\leq \frac{3\sqrt 2}{2}$. However, letting $u=\sin \angle OSR, v=\sin\angle OPS, w=\sin \angle ORQ$, observe that $\tan \angle OSR \tan \angle OPS \tan \angle ORQ=1$. This implies that $uvw=\sqrt{(1-u^2)(1-v^2)(1-w^2)}$ by Pythagorean identity and moving the cosines to the RHS. Now, squaring gives the result sqing wanted us to prove, that is $a,b,c$ are positive real numbers such $(1-a)(1-b)(1-c)=abc$, and we want $1<\sqrt{a}+\sqrt{b}+\sqrt{c}\leq 3\frac{\sqrt 2}{2}$. I used calculus to finish, I might writeup it here later

SPLENDID IDEAS!