APMO 2007

[Ashfaq Uday]

Given \[\sqrt{x}+\sqrt{y}+\sqrt{z}=1\] for all positive real \[x,y,z\]Prove \[\frac{x^2+yz}{\sqrt{2x^2(y+z)}} + \frac{y^2+zx}{\sqrt{2y^2(z+x)}} + \frac{z^2+xy}{\sqrt{2z^2(x+y)}} \geq1\]

[*Mahi*]

$Q.M. \geq A.M.$ gives us $\sqrt {\frac {y+z} 2} \geq \frac {\sqrt y+\sqrt z} 2$
Or, $\sqrt 2 \sqrt{({y+z})} \geq {\sqrt y+\sqrt z}$
And thus $\frac 1 {\sqrt 2 \sqrt{({y+z})}} \leq \frac 1 {\sqrt y+\sqrt z} $

By $Q.M. \geq A.M.$ , \[\sum_{x,y,z} \frac{x^2+yz}{x \sqrt 2 \sqrt {y+z}}\geq\sum_{x,y,z} \frac{x+\frac{yz}{x}}{ \sqrt y +\sqrt {z}}\]

Not so fast!

[*Mahi*]

Great joke... you didn't read the post well did you? If you did then you would have seen that the real inequality and the one you derived has opposite signs, $\leq$ and $\geq$, making your solution incorrect. That's why I added "Not so fast" , because skipping those "innocent" lines can make the proof crumble.
And we all know what the mathlinks people do, they just solves the problem instantly in their head and write it down as fast as their internet works!

[sourav das]

Sorry...... I really did a great mistake....

[sourav das]

\[\sum \frac{x^2+yz}{\sqrt{2x^2(y+z)}}\geq 1\Leftrightarrow 2(\sum \frac{x}{\sqrt{2(y+z)}}) + 2(\sum \frac{yz}{\sqrt{2x^2(y+z)}})\geq 2\].......(*)
W.L.O.G. Let
\[x\geq y\geq z\Rightarrow xy(x-y)+z(x-y)(x+y)\geq 0\Rightarrow x^2(y+z)\geq y^2(z+x)\]
Same way, $y^2(z+x)\geq z^2(x+y)$ And
\[xy\geq zx\geq yz\]
Using re-arrangement inequality and QM$\geq$A.M.
\[2\sum \frac{yz}{\sqrt{2x^2(y+z)}}\geq \sum \frac{y+z}{\sqrt{2(y+z)}}= \sum \sqrt{\frac{y+z}{2}}\geq \sum \frac{\sqrt y + \sqrt z}{2}= 1\].....(i)
\[2\sum \frac{x}{\sqrt{2y+2z}}\geq\sum \frac{y+z}{\sqrt{2y+2z}}= \sqrt{\frac{y+z}{2}}\geq \sum \frac{\sqrt y+\sqrt{z}}{2} = 1\]...........(ii)
Adding (i) and (ii) we'll get (*)

[Phlembac Adib Hasan]

My Proof :

\[\sum_{cyc}\frac{x^2+yz}{\sqrt{2x^2(y+z)}}\geq 1\]
\[\Rightarrow \frac{1}{\sqrt{2}}\sum_{cyc}\frac{x^2+yz}{x\sqrt{y+z}}\geq 1\]\[\Rightarrow \left ( \sum_{cyc}\frac{x}{\sqrt{y+z}} \right )+\left ( \sum_{cyc}\frac{yz}{x\sqrt{y+z}} \right )\geq \sqrt{2}....(*)\]
Before going through the proof we'll prove some lemmas.
Lemma 1:
$3(x+y+z)\ge 1$
It is a direct consequence of Cauchy applying to the sequences $1,1,1$ and $\sqrt {x},\sqrt {y},\sqrt {z}$.
Lemma 2 :
$\sum_{cyc}\frac {xy}{z}\ge x+y+z$
Proof:From AM-GM,
\[2\left ( \sum_{cyc}\frac {xy}{z} \right )=\sum_{cyc}\frac {xy}{z}+\frac {yz}{x}\geq 2(x+y+z)\]
And also this is an obvious case of Muirhead:\[[1,1,-1]\ge [1,0,0]\]
Lemma 3 :
$\frac {1}{3(xy+yz+zx)}\ge \frac {1}{(x+y+z)^2}$
It is a direct consequence of AM-GM.So left for you.

Now we'll start the proof:
From Holder's inequality,\[\left ( \sum_{cyc}\frac{x}{\sqrt{y+z}} \right )\left ( \sum_{cyc}\frac{x}{\sqrt{y+z}} \right )\left ( \sum _{cyc} x(y+z)\right )\geq (x+y+z)^3\]
\[\Rightarrow 2\left ( \sum_{cyc}\frac{x}{\sqrt{y+z}} \right )^2\geq \frac{(x+y+z)^3}{xy+yz+zx}\]
\[\Rightarrow 2\left ( \sum_{cyc}\frac{x}{\sqrt{y+z}} \right )^2\geq \frac{3(x+y+z)^3}{3(xy+yz+zx)}\geq \frac{3(x+y+z)^3}{(x+y+z)^2}\]
\[= 3(x+y+z)\geq 1\]\[\Rightarrow 2\left ( \sum_{cyc}\frac{x}{\sqrt{y+z}} \right )^2\geq 1\]\[\Rightarrow \sum_{cyc}\frac{x}{\sqrt{y+z}} \geq \frac{1}{\sqrt{2}}.....(i)\]

Again from Holder,\[\left ( \sum_{cyc}\frac{yz}{x\sqrt{y+z}} \right )\left ( \sum_{cyc}\frac{yz}{x\sqrt{y+z}} \right )\left ( \sum_{cyc}yzx^2(y+z) \right )\geq \left ( xy+yz+zx \right )^3\]\[\Rightarrow 2\left ( \sum_{cyc}\frac{yz}{x\sqrt{y+z}} \right )^2\geq \frac{\left ( xy+yz+zx \right )^3}{xyz(xy+yz+zx)}=\frac{\left ( xy+yz+zx \right )^2}{xyz}\]\[=\frac{x^2y^2+y^2z^2+z^2x ^2}{xyz}+2\frac{xy^2z+yz^2x+zx ^2y}{xyz}\]\[=\sum_{cyc}\frac {xy}{z}+2(x+y+z)\geq 3(x+y+z)\geq 1\]\[\Rightarrow 2\left (\sum_{cyc}\frac{yz}{x\sqrt{y+z}} \right )^2\geq 1\]\[\Rightarrow \sum_{cyc}\frac{yz}{x\sqrt{y+z}}\geq \frac{1}{\sqrt{2}}....(ii)\]
Adding $(i)$ and $(ii)$ we get $(*)$.\[\left ( \sum_{cyc}\frac{x}{\sqrt{y+z}} \right )+\left ( \sum_{cyc}\frac{yz}{x\sqrt{y+z}} \right )\geq \frac {1}{\sqrt{2}}+\frac {1}{\sqrt{2}}=\sqrt {2}\]
At last we are doooonnnne.