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By what is also known as Titu's form of Cauchy-Schwarz we have c−1a2+d−1b2+e−1c2+a−1d2+b−1e2≥∑a−5(∑a)2≥20, since it comes to (∑a−10)2≥0 . Equality occurs for a=b=c=d=e=2
Method 2:
We know that (a−2)2≥0 so a2≥4(a−1). Similarly we get c−1a2+d−1b2+e−1c2+a−1d2+b−1e2≥c−14(a−1)+d−14(b−1)+e−14(c−1)+a−14(d−1)+b−14(e−1)≥55c−14(a−1)⋅d−14(b−1)⋅e−14(c−1)⋅a−14(d−1)⋅b−14(e−1)=20.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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Comments
@Rajdeep Dhingra
You could tell me those problems and I may be able to help you ? See this
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http://olympiads.hbcse.tifr.res.in/uploads/crmo-2013-paper-4 here question number 3
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Method 1:
By what is also known as Titu's form of Cauchy-Schwarz we have
c−1a2+d−1b2+e−1c2+a−1d2+b−1e2≥∑a−5(∑a)2≥20, since it comes to (∑a−10)2≥0 . Equality occurs for a=b=c=d=e=2
Method 2:
We know that (a−2)2≥0 so a2≥4(a−1). Similarly we get c−1a2+d−1b2+e−1c2+a−1d2+b−1e2≥c−14(a−1)+d−14(b−1)+e−14(c−1)+a−14(d−1)+b−14(e−1)≥55c−14(a−1)⋅d−14(b−1)⋅e−14(c−1)⋅a−14(d−1)⋅b−14(e−1)=20.
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add me also in ur hangouts ... I am also preparing for RMO! I'll give you a hand in solving the problems !!
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Give me your Email ID.
[email protected]
thanks!
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Which class do ya study in ?
u can use any of the following books 1 . arthur engel 2. pre college mathematics 3 . rmo and inmo prep booklet by rajeev manocha
Can you guys help me in solving my note named :"Primes filled with primes".