APMO 2007
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Given \[\sqrt{x}+\sqrt{y}+\sqrt{z}=1\] for all positive real \[x,y,z\]Prove \[\frac{x^2+yz}{\sqrt{2x^2(y+z)}} + \frac{y^2+zx}{\sqrt{2y^2(z+x)}} + \frac{z^2+xy}{\sqrt{2z^2(x+y)}} \geq1\]
Re: APMO 2007
$Q.M. \geq A.M.$ 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} $
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} $
Not so fast!sourav das wrote:By $Q.M. \geq A.M.$ , \[\sum_{x,y,z} \frac{x^2+yz}{x \sqrt 2 \sqrt {y+z}}\geq\sum_{x,y,z} \frac{x+\frac{yz}{x}}{ \sqrt y +\sqrt {z}}\]
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Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Re: APMO 2007
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!
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!
Please read Forum Guide and Rules before you post.
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
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Re: APMO 2007
Sorry...... I really did a great mistake....
You spin my head right round right round,
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
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Re: APMO 2007
\[\sum \frac{x^2+yz}{\sqrt{2x^2(y+z)}}\geq 1\Leftrightarrow 2(\sum \frac{x}{\sqrt{2(y+z)}}) + 2(\sum \frac{yz}{\sqrt{2x^2(y+z)}})\geq 2\].......(*)
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 (*)
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 (*)
You spin my head right round right round,
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
- Phlembac Adib Hasan
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