help ! help ! help !
- nafistiham
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এই অসমতাটা অনেকক্ষণ চেষ্টা করেও পারছি না। যদিও এটা কোন exercise না। তবুও, কিছু exercise এ এটাকে কাজে লাগিয়েছি । একটু হিন্ট দিন
$x^{3} + y^{3} + z^{3} + 3xyz \geq x^{2}(y+z) + y^{2}(x+z) + z^{2}(x+y)$
$x^{3} + y^{3} + z^{3} + 3xyz \geq x^{2}(y+z) + y^{2}(x+z) + z^{2}(x+y)$
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Re: help ! help ! help !
Is this about ex 1.7.3 (new book)?
Then try these
Then try these
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: help ! help ! help !
Mahi র hint এ কাজ হয়ে যাবার কথা। তারপরও না হলে
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......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
- nafistiham
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Re: help ! help ! help !
সর্বনাশ!! বইয়ে যে সরাসরি দেওয়া আছে খেয়াল ই করিনি।
এটা তো দেখি imo তেও আসছে। পুরানো বইয়ে সবকিছু নেই তো।
এটা তো দেখি imo তেও আসছে। পুরানো বইয়ে সবকিছু নেই তো।
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
- Nadim Ul Abrar
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Re: help ! help ! help !
Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$
then we have to prove that $(\frac{x-y+z}{y})(\frac{x-z+y}{z})(\frac{y-x+z}{x}) \leq 1
$
or $(x-y+z)(x+y-z)(y+z-x) \leq xyz
$
after simplifying
we get that we have to prove that
if $x,y,z>0$ then ,
$x^{2}(y+z)+y^{2}(x+z)+z^{2}(x+y) \leq x^{3}+y^{3}+z^{3}+3xyz $
eheheheheh modified version of IMO 2000-2
then we have to prove that $(\frac{x-y+z}{y})(\frac{x-z+y}{z})(\frac{y-x+z}{x}) \leq 1
$
or $(x-y+z)(x+y-z)(y+z-x) \leq xyz
$
after simplifying
we get that we have to prove that
if $x,y,z>0$ then ,
$x^{2}(y+z)+y^{2}(x+z)+z^{2}(x+y) \leq x^{3}+y^{3}+z^{3}+3xyz $
eheheheheh modified version of IMO 2000-2
$\frac{1}{0}$
- nafistiham
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Re: help ! help ! help !
Exercise 1.118. (IMO, 1964) Let a, b, c be positive real numbers. Prove that
\[x^{2}(y+z)+y^{2}(x+z)+z^{2}(x+y) \leq x^{3}+y^{3}+z^{3}+3xyz\]
৩৬ বছর আগে সরাসরি আসছে।
\[x^{2}(y+z)+y^{2}(x+z)+z^{2}(x+y) \leq x^{3}+y^{3}+z^{3}+3xyz\]
৩৬ বছর আগে সরাসরি আসছে।
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
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Re: help ! help ! help !
একটা সরাসরি প্রমান দিছি।
WLOG $a \geq b \geq c$
Now, just re-arrang it like bellow
$a(a-b)(a-c) +b(b-c)(b-a)+ c(c-a)(c-b) \geq 0$
Now,
$c(c-a)(c-b) \geq 0$
$(a-b)(a^2-ac -b^2+bc)=(a-b)((a+b)(a-b) - c(a-b))$
But $a+b\geq c$
So $(a+b)(a-b) - c(a-b) \geq 0$
$(a-b)((a+b)(a-b) - c(a-b)) \geq 0$
Our proof is done.
(Sorry, i proof it for a,b,c )
WLOG $a \geq b \geq c$
Now, just re-arrang it like bellow
$a(a-b)(a-c) +b(b-c)(b-a)+ c(c-a)(c-b) \geq 0$
Now,
$c(c-a)(c-b) \geq 0$
$(a-b)(a^2-ac -b^2+bc)=(a-b)((a+b)(a-b) - c(a-b))$
But $a+b\geq c$
So $(a+b)(a-b) - c(a-b) \geq 0$
$(a-b)((a+b)(a-b) - c(a-b)) \geq 0$
Our proof is done.
(Sorry, i proof it for a,b,c )
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......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
- Nadim Ul Abrar
- Posts:244
- Joined:Sat May 07, 2011 12:36 pm
- Location:B.A.R.D , kotbari , Comilla
Re: help ! help ! help !
oh shourav da .. josss
$x^{3}-x^{2}z-x^{2}y+xyz=x(x-y)(x-z)$
$y^{3}-y^{2}x-y^{2}z+xyz=y(y-x)(y-z)$
$z^{3}-z^{2}x-z^{2}y+xyz=z(z-y)(z-x)
$
put n=1 in scaur's Inequality ...
$x(x-y)(x-z)+y(y-x)(y-z)+z(z-x)(z-y)
\geq0 $
so Its proved ...
That means IMO 2000-02 is solved too
$x^{3}-x^{2}z-x^{2}y+xyz=x(x-y)(x-z)$
$y^{3}-y^{2}x-y^{2}z+xyz=y(y-x)(y-z)$
$z^{3}-z^{2}x-z^{2}y+xyz=z(z-y)(z-x)
$
put n=1 in scaur's Inequality ...
$x(x-y)(x-z)+y(y-x)(y-z)+z(z-x)(z-y)
\geq0 $
so Its proved ...
That means IMO 2000-02 is solved too
$\frac{1}{0}$
Re: help ! help ! help !
কেন যে ৩৬ বছর আগে জন্মাইলাম না ?
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........