shobber wrote: $x$, $y$ and $z$ are positive real numbers such that $x+y+z=1$, prove: \[ \frac{xy}{\sqrt{xy+yz}}+\frac{yz}{\sqrt{yz+zx}}+\frac{zx}{\sqrt{zx+xy}}\leq \frac{\sqrt{2}}{2} \] See here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88439

\[\sum_{cyc}\frac{xy}{\sqrt{xy+yz}}\leq\sum_{cyc}\frac{3x^{2}+2y^{2}+3z^{2}-4yz+20xy}{8\sqrt{2}(x+y+z)}=\frac{x+y+z}{\sqrt{2}}=\frac{1}{\sqrt{2}}.\]

This is actually a China TST problem, it has been posted allready, a classic solution above of the topic is posted by me. http://www.mathlinks.ro/Forum/viewtopic.php?t=124977

chien than wrote: Let $x;y;z>0$ and satisfying $x+y+z=1$ . Prove that: $\frac{xy}{\sqrt{xy+yz}}+\frac{yz}{\sqrt{yz+zx}}+\frac{zx}{\sqrt{zx+xy}}\le \frac{\sqrt 2}2$ By Cauchy Schwarz inequality, we have \[\sum\frac{xy}{\sqrt{xy+yz}}=\sum\frac{x\sqrt{y}}{\sqrt{z+x}}\le \sum\frac{x\sqrt{2y}}{\sqrt{z}+\sqrt{x}}\] Hence, it suffices to prove \[\sum\frac{x\sqrt{y}}{\sqrt{z}+\sqrt{x}}\le \frac{1}{2}\] Setting $a=\sqrt{x},b=\sqrt{y},c=\sqrt{z}$, we have to prove \[\sum\frac{a^{2}b}{c+a}\le \frac{1}{2}(a^{2}+b^{2}+c^{2}) \] or equivalently \[\sum a^{2}+2abc\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right) \ge 2(ab+bc+ca) \] By AM-GM inequality, we have \[2abc\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right) \ge \frac{9abc}{a+b+c}\] Hence, it suffices to prove \[\sum a^{2}+\frac{9abc}{a+b+c}\ge 2\sum ab \] which is Schur inequality. We are done. PS:this is China 2005, not yours, Chienthan.

very nice solution, toan!

I see that it is not very strong ineq!

Math pro wrote: I see that it is not very strong ineq! No, not really actually^^ shobber wrote: Given three positive real numbers $ x$, $ y$, $ z$ such that $ x + y + z = 1$, prove that \[ \frac {xy}{\sqrt {xy + yz}} + \frac {yz}{\sqrt {yz + zx}} + \frac {zx}{\sqrt {zx + xy}} \le \frac {\sqrt {2}}{2}.\] My solution goes as follows: We have \[ \sum \frac {xy}{\sqrt {xy + yz}} = 2 \sum \frac {x + y}{2} \sqrt {\frac {x^2y}{(x + y)^2(x + z)}} = S.\] Since the function $ f(x) = \sqrt {x}$ is concave on $ \mathbb{R}_ +$, we have by Jensen \[ S \le 2 \sqrt {\sum \frac {x^2y}{2(x + y)(x + z)}} = \sqrt {2}\sqrt {\frac {\sum (x^2y^2 + x^2yz)}{(x + y)(y + z)(z + x)}}.\] Thus, after squaring, it remains to prove $ \sum xy(x - y)^2 \ge 0,$ which is obvious.

we have that \[ \sum_{cyc} \frac{xy}{\sqrt{xy+yz}}\le \sqrt{(\sum xy(x+y))(\sum \frac{x}{(x+y)(x+z)})}\le \frac{\sqrt{2}}{2}. \] So, it's equivalent to $ \sum_{cyc} a^2(b+c)(a-b)(a-c)\ge 0$

My solution is I guess tough for no reason Applying jensen's to the function $f(x)=\sqrt x$ $$\frac {xy}{\sqrt {xy + yz}} + \frac {yz}{\sqrt {yz + zx}} + \frac {zx}{\sqrt {zx + xy}} \le \sqrt{\sum_{x,y,z} \frac{2x^2y^2(xy+yz+zx)}{(xy+yz)(xy+xz)}}$$Now we are left to prove $$\sum_{x,y,z} \frac{2x^2y^2(xy+yz+zx)}{(xy+yz)(xy+xz)} \le \frac{1}{2}$$ For this, see http://www.artofproblemsolving.com/community/c6h1160617_weird_symmetric_inequality

shobber wrote: Given three positive real numbers $ x$, $ y$, $ z$ such that $ x + y + z = 1$, prove that $ \frac {xy}{\sqrt {xy + yz}} + \frac {yz}{\sqrt {yz + zx}} + \frac {zx}{\sqrt {zx + xy}} \le \frac {\sqrt {2}}{2}$. By AM-GM,we have \[xy\sqrt{\frac{2}{xy+yz}}=\frac{2x\sqrt{2(xy+yz)(y+z)^2}}{2(x+z)(y+z)}\]\[\le{\frac{x[2(xy+yz)+(y+z)^2]}{2(x+z)(y+z)}}=\frac{x(2xy+4yz+y^2+z^2)}{2(x+z)(y+z)}\]Hence,we need to prove that \[\sum{\frac{x(2xy+4yz+y^2+z^2)}{2(x+z)(y+z)}}\le{x+y+z}\]\[<=>x+y+z-\sum{\frac{x(2xy+4yz+y^2+z^2)}{2(x+z)(y+z)}}=\frac{\sum{zx(x-y)^2}}{2(x+y)(y+z)(z+x)}\ge{0}\]

toanhocmuonm where did you learn schur's inequality?

sasukerikudo wrote: toanhocmuonm where did you learn schur's inequality? Though I know the question is not directed to me, anyways, Secrets in Inequalities by Pham Kim Hung is a wonderful read.

Ji Chen wrote: \[\sum_{cyc}\frac{xy}{\sqrt{xy+yz}}\leq\sum_{cyc}\frac{3x^{2}+2y^{2}+3z^{2}-4yz+20xy}{8\sqrt{2}(x+y+z)}\] Can anyone tell me why it is true?

Use BW. You'll get something obvious. For example, for $x=\min\{x,y,z\}$, $y=x+u$ and $z=x+v$ we need to prove that: $$24(4u^2-12uv+11v^2)x^3+4(8u^3-16u^2v-8uv^2+43v^3)x^2+$$$$+8(u^4+48u^3v-96u^2v^2+56uv^3+30v^4)x+(2u^2-4uv+3v^2)^2v\geq0,$$which is obvious.

arqady wrote: Use BW. You'll get something obvious. For example, for $x=\min\{x,y,z\}$, $y=x+u$ and $z=x+v$ we need to prove that: $$24(4u^2-12uv+11v^2)x^3+4(8u^3-16u^2v-8uv^2+43v^3)x^2+$$$$+8(u^4+48u^3v-96u^2v^2+56uv^3+30v^4)x+(2u^2-4uv+3v^2)^2v\geq0,$$which is obvious. Thanks

\begin{align*} \text{Observe: } \\ \frac {xy}{\sqrt {xy + yz}} + \frac {yz}{\sqrt {yz + zx}} + \frac {zx}{\sqrt {zx + xy}} \leq \frac {\sqrt {2}}{2}\\ \iff \sum_{\text{cyc}} x\sqrt{\frac{y}{x+z}} \leq \frac{\sqrt{2}}{2}\\ \iff \sum_{\text{cyc}} \frac{x+y}{2} \sqrt{\frac{x^2y}{(x+z)(x+y)^2}} \leq \frac{\sqrt{2}}{2}. \\ \text{Weighted Jensen: }\\ \sum_{\text{cyc}} \frac{x+y}{2} \sqrt{\frac{4x^2y}{(x+z)(x+y)^2}} \\ \leq \sum_{\text{cyc}} \frac{2x^2y}{(x+y)(x+y)}.\\ \text{Notice: }\\ \sum_{\text{cyc}} \frac{x^2y}{(x+y)(x+z)} \leq \frac{1}{4} \\ \iff 4\sum_{\text{cyc}} x^2y(y+z) \leq (x+y+z) \prod_{\text{cyc}} (x+y). \\ \textbf{Q.E.D.} \end{align*}

Using Cauchy Schwarz inequality, $LHS^2=\left(\sum\limits_{cyc}\dfrac{x\sqrt y}{\sqrt {x}+\sqrt z}\right)^2\leqslant\sum\limits_{cyc}\dfrac{x}{x+z}\sum\limits_{cyc}xy.$ $\therefore$ We have to prove $\sum\limits_{cyc}\dfrac{x}{x+z}\sum\limits_{cyc}xy\leqslant\frac{1}{2}\Leftrightarrow 2\sum\limits_{cyc}\dfrac{x}{x+z}\sum\limits_{cyc}xy\leqslant\left(\sum\limits_{cyc}x\right)^2.$ $2\sum\limits_{cyc}\dfrac{x}{x+z}\sum\limits_{cyc}xy=2\sum\limits_{cyc}\dfrac{x}{x+z}(xy+yz)+2\sum\limits_{cyc}\dfrac{x}{x+z}\cdot zx=2\sum\limits_{cyc}xy+2\sum\limits_{cyc}\dfrac{x^2z}{x+z}$ $\leqslant 2\sum\limits_{cyc}xy+2\sum\limits_{cyc}\dfrac{x^2z}{2\sqrt{xz}}=2\sum\limits_{cyc}xy+\sum\limits_{cyc}x^{\frac{3}{2}}z^{\frac{1}{2}}\leqslant2\sum\limits_{cyc}xy+\sum\limits_{cyc}x^2=\left(\sum\limits_{cyc}x\right)^2.\Box$