shobber wrote: Let $x,y,z$ be positive numbers such that \[ {1\over x}+{1\over y}+{1\over z}=1. \] Show that \[ \sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z} \] From the condition we have $\sum xy= xyz$ $S=\sum \sqrt{x+yz} \geq \sqrt{(\sum {\sqrt{x}})^2+(\sum \sqrt{yz})^2}$ Put $u=\sum \sqrt{x}$ $v=\sum \sqrt{yz}$ $\sqrt{u^2+v^2} \geq \sqrt{xyz}+u$ $u^2+v^2 \geq xyz +u^2+2\sum \sqrt{x}\sqrt{xyz}$ $\sum xy + 2\sqrt{xyz}\sum \sqrt{x} \geq xyz +2\sqrt{xyz} \sum \sqrt{x}$ But we have $\sum xy =xyz$

Isn't that just walking forwards and going backwards?

Aren't those equivalences, Karth? That would make the proof valid. shobber wrote: Let $x,y,z$ be positive numbers such that \[{1\over x}+{1\over y}+{1\over z}=1.\] Show that \[\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}\] Let $\sum = \sum_{\textrm{cyclic}}$. Square both sides to get the equivalent \[\sum (x+yz)+2\sum\sqrt{(x+yz)(y+zx)}\geq (xyz+x+y+z)+2\left(\sum (x+1)\sqrt{yz}\right)\] It's given that $xy+yz+zx=xyz$, so we just want to prove that \[\sum\sqrt{(x+yz)(y+zx)}\geq \sum (x+1)\sqrt{yz}\] But if we could prove that $\sqrt{(x+yz)(y+zx)}\geq (z+1)\sqrt{xy}$, we would be done. This inequality is equivalent with $xy+x^{2}z+y^{2}z+xyz^{2}\geq xyz^{2}+2xyz+xy$, i.e. $z(x-y)^{2}\geq 0$, which is obvious.

shobber wrote: Let $x,y,z$ be positive numbers such that \[{1\over x}+{1\over y}+{1\over z}=1.\] Show that \[\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}\] $\sum \sqrt{x+yz}= \sum \sqrt{x \frac{xyz}{xy+yz+zx}+yz}$ $=\sum \sqrt{\frac{yz}{xy+yz+zx}(x^{2}+xy+yz+zx)}$ $=\sum \sqrt{\frac{yz(x+y)(x+z)}{xy+yz+zx}}$ $=\frac{1}{\sqrt{xy+yz+zx}}\sum \sqrt{yz(x+y)(x+z)}$ $\geq \frac{1}{\sqrt{xy+yz+zx}}\sum \sqrt{yz(x+\sqrt{yz})^{2}}$ $= \frac{1}{\sqrt{xy+yz+zx}}\sum (x\sqrt{yz}+yz)$ $=\sum (\sqrt{\frac{x^{2}yz}{xy+yz+zx}}+\frac{yz}{\sqrt{xy+yz+zx}})$ $=\sum \sqrt{x}+\frac{\sum{yz}}{\sqrt{xy+yz+zx}}$ $=\sum \sqrt{x}+\sqrt{xy+yz+zx}$ $=\sum \sqrt{x}+\sqrt{xyz}$.

The problem duces to $a+b+c=1$, and a problem solved by Minkowski ineq.

Let $a=\frac{1}{x}$, $b=\frac{1}{y}$, $c=\frac{1}{z}$. Then, $a+b+c=1$ $\sum = \sum_{cyc}$ $LHS = \sum \sqrt{x+yz}= \sum \frac{\sqrt{bc+a}}{\sqrt{abc}}= \sum \frac{bc+a(a+b+c)}{\sqrt{abc}}= \sum \frac{\sqrt{(a+b)(a+c)}}{\sqrt{abc}}\geq \sum \frac{a+\sqrt{bc}}{\sqrt{abc}}= \sum \sqrt{xyz}(\frac{1}{x}+\frac{1}{\sqrt{yz}}) = \sqrt{xyz}\sum \frac{1}{x}+\sum \sqrt{x}= \sqrt{xyz}+\sum \sqrt{x}= RHS$ Is it the same as your Minkowski ineq? If yes then I am sorry.

Does this work? PS sorry for bringing up an old topic. Solution By Cauchy-Schwarz, it follows that for any $ x,y,z > 0$, $ (x + yz)(y + xz) \ge (\sqrt {xy} + z \sqrt {xy})^2$ $ \sqrt {(x + yz)(y + xz)} \ge \sqrt {xy} + z \sqrt {xy}$ Summing this for all pairings of $ x,y,z$, it follows that $ \sqrt {(x + yz)(y + xz)} + \sqrt {(x + yz)(z + xy)} + \sqrt {(z + xy)(y + xz)} \ge \sqrt {xy} + \sqrt {xz} + \sqrt {yz} + z \sqrt {xy} + x \sqrt {yz} + y \sqrt {xz}$ Now note that by the condition, $ xy + xz + yz = xyz$. Multiplying by $ 2$ and adding $ x + y + z + xy + xz + yz = x + y + z + xyz$ gives that, $ (\sqrt {x + yz} + \sqrt {y + zx} + \sqrt {z + xy})^2 \ge (\sqrt {xyz} + \sqrt {x} + \sqrt {y} + \sqrt {z})^2$ The result follows.

$ \sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}$ $ \iff \sum_{cyc} \sqrt{\frac{1}{yz}+\frac{1}{x}} \ge 1+ \sum_{cyc} \frac{1}{\sqrt{yz}}$ $ \iff \sum_{cyc} \sqrt{(\frac{1}{x}+\frac{1}{y})(\frac{1}{x}+\frac{1}{z})} \ge 1+ \sum_{cyc} \frac{1}{\sqrt{yz}}$ (since $ (\frac{1}{x}+\frac{1}{y})(\frac{1}{x}+\frac{1}{z})=\frac{1}{yz}+\frac{1}{x}$) By C-S inequality, $ \sum_{cyc} \sqrt{(\frac{1}{x}+\frac{1}{y})(\frac{1}{x}+\frac{1}{z})} \ge \sum_{cyc}(\frac{1}{x}+\frac{1}{\sqrt{yz}} )=1+ \sum_{cyc} \frac{1}{\sqrt{yz}}$ ( since $ {1\over x}+{1\over y}+{1\over z}=1$) QED

After dividing by $\sqrt{xyz}$ and making the substitution $x=\frac{1}{a}, y=\frac{1}{b}$ and $z=\frac{1}{c}$, the inequality becomes \[ \sqrt{a+bc} +\sqrt{b+ca} +\sqrt{c+ab}\geqslant 1+\sqrt{ab}+\sqrt{bc}+\sqrt{ca} \] Notice that $a+bc=(a+b)(a+c)$ and by Cauchy Schwarz $(a+b)(a+c)\geqslant (a+\sqrt{bc})^2$. Using this we have \[ \sqrt{a+bc} +\sqrt{b+ca} +\sqrt{c+ab}\geqslant a+\sqrt{bc} +b+ \sqrt{ca} +c +\sqrt{ab} \] We're done due to $a+b+c=1$.

The condition gives \begin{align*} xyz = xy + xz + yz \iff yz + x = x + y + z + \frac {yz} {x} \end{align*}Furthermore, $\sqrt {xyz} = \left (\dfrac {1} {x} + \dfrac {1} {y} + \dfrac {1} {z}\right) \sqrt {xyz} = \sqrt {\dfrac {xy} {z}} + \sqrt {\dfrac {xz} {y}} + \sqrt {\dfrac {yz} {x}}$, so it suffices to show \begin{align*} \sum \sqrt {\dfrac {yz} {x} + x + y + z} \ge \sum \sqrt {\dfrac {xy} {z}} + \sum \sqrt {x} \end{align*}From AM-GM, \[\dfrac {xy} {z} + x + y + z \ge \dfrac {xy} {z} + z + 2 \sqrt {xy} \implies \sqrt {\dfrac {xy} {z} + x + y + z} \ge \sqrt {\dfrac {xy} {z}} + \sqrt {z}\]Summing cyclically gives the desired result.

It's equivalent to show \[ \sum_{\text{cyc}} \sqrt{x+yz\left( \frac1x+\frac1y+\frac1z \right)} \ge \sum_{\text{cyc}} \sqrt{x} + \sqrt{xyz} \cdot \frac1x \] but now we're done by Cauchy as \[ \sqrt{x+yz\left( \frac1x+\frac1z+\frac1z \right)}= \frac{\sqrt{(x+y)(x+z)}}{\sqrt{x}} \ge \frac{x+\sqrt{yz}}{\sqrt{x}} = \sqrt x + \frac{\sqrt{xyz}}{x}. \]

really a nice one.

The inequality is equivalent to Let $a, b ,c$ be positive real numbers such that $a+b+c=1$ , prove that\[ \frac{a}{\sqrt{a+bc}+\sqrt{bc}}+\frac{b}{\sqrt{b+ca}+\sqrt{ca}}+\frac{c}{\sqrt{c+ab}+\sqrt{ab}}\ge 1.\] vyfukas wrote: We have that $b+c=1-a \Rightarrow 1-2a+a^2=b^2+2bc+c^2\ge 4bc \Rightarrow bc \le \dfrac{1-2a+a^2}{4}$ Then we substitute this result in a given inequality: $\dfrac{a}{\sqrt{a+bc}+\sqrt{bc}}\ge\dfrac{a}{\sqrt{a+\dfrac{1-2a+a^2}{4}}+\sqrt{\dfrac{1-2a+a^2}{4}}}=\dfrac{a}{\dfrac{a+1}{2}+\dfrac{1-a}{2}}=a$ Then adding similar inequalities we will get $a+b+c=1$ thus we are done. http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=568982

Since $xyz=xy+yz+zx$ we see that the individual terms on the RHS and LHS are equal when squared.So after squaring the ineq,it is sufficient to prove $$\sum \sqrt{(x+yz)(y+xz)} \ge \sum \sqrt{xy} +\sum z\sqrt{xy}$$which is Cauchy schwarz in 'most normal form'

kapilpavase wrote: Since $xyz=xy+yz+zx$ we see that the individual terms on the RHS and LHS are equal when squared.So after squaring the ineq,it is sufficient to prove $$\sum \sqrt{(x+yz)(y+xz)} \ge \sum \sqrt{xy} +\sum z\sqrt{xy}$$which is Cauchy schwarz in 'most normal form' I am afraid, your solution is not different enough from post #8 to bring up such an old topic.

solution by tuchuyenhg.

shobber wrote: Let $x,y,z$ be positive numbers such that ${1\over x}+{1\over y}+{1\over z}=1.$ Show that \[ \sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z} \] http://www.artofproblemsolving.com/community/c6h568982p3338102 Let $x,y,z$ be positive numbers such that $x+ y+ z=1.$ Show that \[ \sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z} \]Stronger than APMO 2002

http://www.artofproblemsolving.com/community/c6h133953p758077 If $ x$, $ y$, $ z$ are three nonnegative reals, then prove that$$ \sum_{\text{cyc}}\sqrt {\left(z + x\right)\left(x + y\right)}\geq x + y + z + \sqrt3\cdot\sqrt {yz + zx + xy}$$http://www.artofproblemsolving.com/community/c6h200514p1102800 Prove if $a,b,c>0$ and $ab+bc+ca=3$ then $$ \sum\sqrt{a^2+3}\geq a+b+c+3$$

Homogenise the inequality to $$ \sum_\text{cyc} \sqrt{\frac{(x+y)(x+z)}{x}} \ge \sum_\text{cyc} \sqrt{x} + \sqrt{\frac{yz}{x}}$$We show that $\sqrt{\frac{(x+y)(x+z)}{x}} \ge \sqrt{x} + \sqrt{\frac{yz}{x}}$. Upon squaring and expanding, it reduces to $y+z \ge 2\sqrt{yz}$ which is obvious by AM-GM.

WizardMath wrote: We show that $\sqrt{\frac{(x+y)(x+z)}{x}} \ge \sqrt{x} + \sqrt{\frac{yz}{x}}$. This is true by Cauchy, no need to square both the sides

Yeah very true. Wizard math nice. Solution. In inequality expansion and getting is great

Karamata inequality can kill this problem immediately.

Let $ a,b,c>0 $ and $ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1 $ . Prove that $$\sqrt{\frac{a-1}{b}}+\sqrt{\frac{b-1}{c}}+\sqrt{\frac{c-1}{a}} \ge \sqrt{6}$$$$\iff$$$$a\sqrt{1+\frac{c}{b}}+b\sqrt{1+\frac{a}{c}}+c\sqrt{1+\frac{b}{a}} \ge \sqrt{6(ab+bc+ca)}$$h

Wanted to prove by Am-Gm

a different solution: $$\sum \sqrt{x+yz} \geq \sqrt{xyz} +\sum \sqrt{x} \Leftrightarrow \sum \frac{yz}{\sqrt{x+yz}+\sqrt x} =\sum \sqrt{x+yz}-\sqrt{x}\geq\sqrt{xyz}$$which is correct because: $$\sum \frac{yz}{\sqrt{x+yz}+\sqrt x}=\sum \frac{yz}{\sqrt{x+yz(\frac 1x +\frac 1y +\frac 1z)}+\sqrt x}=\sqrt{xyz}\sum \frac{yz}{\sqrt{(xz+yz)(xy+zy)}+ x\sqrt{yz}}\geq \sqrt{xyz}\sum \frac{yz}{\frac{xz+yz +xy +yz}{2} +\frac{xy+xz}{2}}=\sqrt{xyz}$$

Let $x,y,z$ be positive numbers such that $\frac{1}{x}+\frac{2}{yz}=1.$ Show that $$\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\geq\frac{3\sqrt{3}}{5}(\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z} )$$

Note that $\sqrt{xyz} = \sqrt{\frac{yz}{x}} + \sqrt{\frac{zx}{y}} + \sqrt{\frac{xy}{z}}$. Now Note that we need to prove $\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy} \ge \sqrt{\frac{yz}{x}} + \sqrt{\frac{zx}{y}} + \sqrt{\frac{xy}{z}}+\sqrt{x}+\sqrt{y} + \sqrt{z}$ so we will prove $\sqrt{x+yz} \ge \sqrt{\frac{yz}{x}} + \sqrt{x}$ or $x + yz \ge \frac{yz}{x} + x + 2\sqrt{yz}$. Note that $\frac{yz+zx+xy}{xyz} = 1 \implies yz = xyz - zx- xy \implies \frac{yz}{x} = zy - z - y$ so we need to prove $y + z \ge 2\sqrt{yz}$ which is obviously true.

The condition is equivalent to $xyz=xy+yz+zx$. We'll use this to homogenize the left-hand side of the inequality: \[\sqrt{xyz}\sum\limits_{sym} \sqrt{x+yz}=\sum\limits_{sym} \sqrt{x\cdot xyz+yz\cdot (xy+yz+zx)}=\sum\limits_{sym}\sqrt{yz}\cdot \sqrt{(x+y)(x+z)}\]Now note that using the Cauchy-Schwarz inequality, we have that: \[\sqrt{xyz}\sum\limits_{sym} \sqrt{x+yz}=\sum\limits_{sym}\sqrt{yz}\cdot \sqrt{(x+y)(x+z)}\geq \sum\limits_{sym}\sqrt{yz}\cdot (x+\sqrt{yz})=\sqrt{xyz}\sum\limits_{sym} \sqrt{x}+xyz\sum\limits_{sym} \frac{1}{x}=\sqrt{xyz}\sum\limits_{sym} \sqrt{x}+xyz\]which is equivalent to the original inequality after diving both sides by $\sqrt{xyz}>0$.

shobber wrote: Let $x,y,z$ be positive numbers such that $ {1\over x}+{1\over y}+{1\over z}=1. $ Show that \[ \sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z} \]

shobber wrote: Let $x,y,z$ be positive numbers such that \[ {1\over x}+{1\over y}+{1\over z}=1. \]Show that \[ \sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z} \] $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ $${1\over x}+{1\over y}+{1\over z}=1$$$$\Rightarrow xy+yz+zx=xyz...(i)$$$$\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}$$$$\Leftrightarrow x+yz+y+zx+z+xy+2\sum_{cyc} \sqrt{xy+x^2z+y^2z+xyz^2}\ge xyz+x+y+z+2\sum_{cyc}\sqrt{x^2yz}+\sqrt{xy}$$By $(i):$ $$\Leftrightarrow \sum_{cyc}\sqrt{(x+yz)(y+xz)}\ge \sum_{cyc}\sqrt{xyz^2}+\sqrt{xy}$$It is enough to prove that: $$\sqrt{(x+yz)(y+xz)}\ge \sqrt{xyz^2}+\sqrt{xy}$$$$\Leftrightarrow xy+x^2z+y^2z+xyz^2\ge xyz^2+xy+2\sqrt{x^2y^2z^2}$$$$\Leftrightarrow (x^2+y^2)\ge 2xy$$which is trivial by $AM-GM$ $$\Rightarrow \sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$