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Here goes first post of 2025! Great blog.
by
math_holmes15, Jan 14, 2025, 8:53 AM
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First post of 2024
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Yiyj1, Feb 8, 2024, 5:40 AM
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First post of 2023
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HoRI_DA_GRe8, Jul 22, 2023, 7:45 AM
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Nice blog ! Your isogonality lemma is really powerful !
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554183, Oct 14, 2021, 8:55 AM
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Post plss....
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samrocksnature, Apr 11, 2021, 10:12 PM
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alas,this is ded
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Hamroldt, Mar 18, 2021, 4:13 PM
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Thanks for the nice blog.
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Feridimo, Mar 6, 2020, 4:17 PM
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I think this might be silly but ... when should we expect to have another post ?? I am very keen to see it
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gamerrk1004, Nov 4, 2019, 4:54 PM
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Let's all echo what's written in the blog description - Stay Insane / 'Cause it's your labor, will and pain/ That takes you to the top of soda fountain
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Kayak, Oct 2, 2017, 7:18 PM
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hey utkarsh jee is over now ... continue your elementary blog pleaseeeeeee!
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kk108, Jun 17, 2017, 11:19 AM
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Congrats on becoming a contest moderator!
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Ankoganit, Mar 9, 2017, 5:22 AM
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INTERSTING BLOG
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kk108, Feb 19, 2017, 2:04 PM
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I have no plans for this blog right now....
No time here people !
But lets see....
I may try some combinatorics 
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utkarshgupta, Feb 15, 2017, 12:47 PM
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Thanks for the nice blog!
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Orkhan-Ashraf_2002, Feb 13, 2017, 6:34 PM
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Revive it!!!
Best blog out there, for sure!
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rmtf1111, Jan 12, 2017, 6:02 PM
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Oh you certainly will..
Bu I don't think I will 
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achilles04, Feb 11, 2015, 6:24 PM
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Thanks
Hope you are there too ...
And hope I get selected
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utkarshgupta, Feb 11, 2015, 11:14 AM
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Nice blog ..
keep it up and you might one day become a mathematician lIke me
( just joking )
Btw don't forget us after going to imotc.
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achilles04, Feb 10, 2015, 4:49 PM
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nice blog!!!!!
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pradhit, Feb 7, 2015, 11:51 AM
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No ideas about the cut off but should be less than 60 according to me...
So you never know
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utkarshgupta, Feb 6, 2015, 4:06 PM
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Hi utkarsh,
what do u expect the cutoff to be this year??
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kgdpsdurg, Feb 6, 2015, 12:41 PM
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Hi utkarsh
How many did you clear in INMO 2015?
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moldevort, Feb 1, 2015, 3:59 PM
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@ Anonymous Bunny
If we both (that should have included me) are lucky enough ! 
by
utkarshgupta, Sep 17, 2014, 2:44 AM
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Nice blog. Great job!
If I am lucky enough, hopefully we shall meet at IMOTC. 
by
AnonymousBunny, Sep 16, 2014, 8:08 PM
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Thanx all !!
If anyone wish to contribute anything just pm !
I will add u to the contributors' list !
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utkarshgupta, Sep 16, 2014, 6:35 AM
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doing good progress!! btw nice blog.
by
rachitgoel, Sep 15, 2014, 4:21 PM
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Nice blog.
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Kezer, Aug 5, 2014, 4:04 PM
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Hi Utkarsh! Wonderful blog. Keep it up.
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YESMAths, Jul 20, 2014, 1:36 PM
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I need some shouts and characters
by
utkarshgupta, Jul 20, 2014, 4:22 AM
by utkarshgupta, Feb 3, 2015, 5:01 PM 
, 
and 
are three real numbers, all different from 
, such that 
, then prove that
.
sign for cyclic summation, this inequality could be rewritten as 
.)![\[\frac {x^{2}}{\left(x - 1\right)^{2}} + \frac {y^{2}}{\left(y - 1\right)^{2}} + \frac {z^{2}}{\left(z - 1\right)^{2}} \geq 1\]](//latex.artofproblemsolving.com/f/a/6/fa6c46c12e7e6bf0b04f6f9a910df6c2b452fe6a.png)
, 
and 
![\[\iff \sum x^2(y^2-2y+1)(z^2-2z+1) \ge (-x-y-z+xyz)^2\]](//latex.artofproblemsolving.com/c/f/3/cf350ec5cbc59afb8e62fea3547cc20005be5ee0.png)
![\[\iff \sum (1+x^2y^2+x^2z^2+x^2-2xy-2xz+2x+2x^2yz-2x^2y-2x^2z) \ge (u-v)^2\]](//latex.artofproblemsolving.com/2/3/4/234503316aa7f67f71a023c88717c7016c6e5e7c.png)
![\[\iff 3+\sum x^2y^2 + \sum x^2 -4\sum xy +2 \sum x + 2\sum x - \sum x^2y+xy^2 \ge (v-u)^2\]](//latex.artofproblemsolving.com/f/2/e/f2eaca00a614ed392778db32255cd4deeeb96985.png)
![\[\iff 3+2(v^2-2uw)+(u^2-2v)-4v + 2u + 2u-\sum x^2y+xy^2 \ge (v-u)^2\]](//latex.artofproblemsolving.com/b/b/f/bbf44220a1cd2b2ebc4d255f20fa2b7a5cbfda53.png)
![\[\iff 3+2v^2-4u+u^2-2v-4v+4u \sum x^2y+xy^2 \ge (v-u)^2\]](//latex.artofproblemsolving.com/e/2/3/e23578b6057409e4eab7c8c37cf6fb5038e0c170.png)
![\[\iff 3+v^2+2uv \ge 6v + 2\sum xy(x+y)\]](//latex.artofproblemsolving.com/4/5/5/455887ff14ef8f87ce4db3cc22edf0a090648f8d.png)
![\[\iff 3+v^2+2uv \ge 6v + 2\sum xy(u-z)\]](//latex.artofproblemsolving.com/0/4/3/0431ec77c5728f40eebdf3fb322895e8b694b247.png)
![\[\iff 3+v^2+2uv \ge 6v + 2uv - 6w\]](//latex.artofproblemsolving.com/5/6/6/566ff9a59117c9854959bc7bc90421ea00a95a4f.png)
![\[\iff (v-3)^2 \ge 0\]](//latex.artofproblemsolving.com/4/9/0/49076d6354d3119b9b03defea75e0608c39fa5d0.png)
,
, 
and 
are three real numbers, all different from 
, such that 
, then prove that
.
sign for cyclic summation, this inequality could be rewritten as 
.)![\[\frac {x^{2}}{\left(x - 1\right)^{2}} + \frac {y^{2}}{\left(y - 1\right)^{2}} + \frac {z^{2}}{\left(z - 1\right)^{2}} \geq 1\]](http://latex.artofproblemsolving.com/f/a/6/fa6c46c12e7e6bf0b04f6f9a910df6c2b452fe6a.png)
, 
and 
![\[\iff \sum x^2(y^2-2y+1)(z^2-2z+1) \ge (-x-y-z+xyz)^2\]](http://latex.artofproblemsolving.com/c/f/3/cf350ec5cbc59afb8e62fea3547cc20005be5ee0.png)
![\[\iff \sum (1+x^2y^2+x^2z^2+x^2-2xy-2xz+2x+2x^2yz-2x^2y-2x^2z) \ge (u-v)^2\]](http://latex.artofproblemsolving.com/2/3/4/234503316aa7f67f71a023c88717c7016c6e5e7c.png)
![\[\iff 3+\sum x^2y^2 + \sum x^2 -4\sum xy +2 \sum x + 2\sum x - \sum x^2y+xy^2 \ge (v-u)^2\]](http://latex.artofproblemsolving.com/f/2/e/f2eaca00a614ed392778db32255cd4deeeb96985.png)
![\[\iff 3+2(v^2-2uw)+(u^2-2v)-4v + 2u + 2u-\sum x^2y+xy^2 \ge (v-u)^2\]](http://latex.artofproblemsolving.com/b/b/f/bbf44220a1cd2b2ebc4d255f20fa2b7a5cbfda53.png)
![\[\iff 3+2v^2-4u+u^2-2v-4v+4u \sum x^2y+xy^2 \ge (v-u)^2\]](http://latex.artofproblemsolving.com/e/2/3/e23578b6057409e4eab7c8c37cf6fb5038e0c170.png)
![\[\iff 3+v^2+2uv \ge 6v + 2\sum xy(x+y)\]](http://latex.artofproblemsolving.com/4/5/5/455887ff14ef8f87ce4db3cc22edf0a090648f8d.png)
![\[\iff 3+v^2+2uv \ge 6v + 2\sum xy(u-z)\]](http://latex.artofproblemsolving.com/0/4/3/0431ec77c5728f40eebdf3fb322895e8b694b247.png)
![\[\iff 3+v^2+2uv \ge 6v + 2uv - 6w\]](http://latex.artofproblemsolving.com/5/6/6/566ff9a59117c9854959bc7bc90421ea00a95a4f.png)
![\[\iff (v-3)^2 \ge 0\]](http://latex.artofproblemsolving.com/4/9/0/49076d6354d3119b9b03defea75e0608c39fa5d0.png)
,
September 2017
February 2017