Find the minimum value of 1 + a 4 + b 8 if a and b are positive real numbers with a b 2 = 5 .
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b 2 = a 5 . b 8 = ( a 5 ) 4 . f ( a ) = 1 + a 4 + ( a 5 ) 4 , f ′ ( a ) = 0 ⟹ a = 5 . f " ( a ) > 0 , s o a = 5 . i s m i n . f ( a ) m i n = 1 + ( 5 ) 4 + ( 5 5 ) 4 = 5 1
Sir can't we apply AM-GM on a^4 and b^8 ?? by applying we get that a^4+b^4=50 . so minimum value = 50+1=51
a and b being positive integers does not satisfy the answer . They must be positive reals.
Thanks for reporting. Fixed!
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Why can a not be 5 and b be 1? Aren't these both integers that make the equation true? Then the answer would be 502
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That's what I had changed in the question that they are real numbers not integers but a moderator had again changed it to integers. Taking them as real numbers you will have the answer as 51 only, just apply AM-GM or Cauchy
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@Satyajit Ghosh – why isnt the ans 51 at { a = 5 b = 4 5 . by simple am-gm?
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@Aareyan Manzoor – I don't understand. The answer is 51 only (what it shows in my pc) btw @Andrew Ellinor sir has already changed my question to integer earlier and it might even happen he changed the answer to something different. Btw what is the new answer? Also the name of my problem has been changed. This is probably a work of a moderator.
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@Satyajit Ghosh – 627 new answer.
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@Aareyan Manzoor – See the reports section, Sir Andrew Ellinor has changed the answer back to 51 and you have probably got your ratings for this question.
@Satyajit Ghosh – I see. Thank you. I am new here and do not know the rules. I just wanted to brush up on my math skills that I have not used in 15 years.
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@Amanda Bloomer Houser – See the reports section, Sir Andrew Ellinor has changed the answer back to 51 and you have probably got your ratings for this question.
Seems extremely awkward the number 145. You should have waited for 150 or maybe you should have infact written 150 as no one is there who will check your profile before attempting the question.
There will be another question for 150. Btw people expect answer here.
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Please change the line of question to a and b are positive reals
They are not positive integers they are positive reals bangali
its direct application of am gm
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That's why there is another question in the sequel which is hopefully more challenging that I made in the sst period xD
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kushagra and lakshya are your classmates??
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@Dev Sharma – Kushagra is but not lakshya. But why? N try my next question
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@Satyajit Ghosh – solved......
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@Dev Sharma – Nice but why did you ask about that question n I'll be also uploading a question for 150 followers problem stay tuned
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I wanted to put emphasis on a common misuse of the AM-GM inequality on 1 + a 4 + b 8 :
Since a, b are positive, 1 + a 4 + b 8 ≥ 3 × 3 a 4 b 8 = 3 ( a b 2 ) 3 4 = 3 × 5 3 4
Equality occurs when 1 = a 4 = b 8 = 5 3 4 . But 1 = 5 3 4 is a contradiction. So while 3 × 5 3 4 represents a lower bound on the expression, it does n o t a minimum.
Instead, apply the AM-GM Inequality on a 4 + b 8 .
a 4 + b 8 ≥ 2 × 2 a 4 b 8 = ( a b 2 ) 2 = 2 × 5 2
Equality occurs when a 4 = b 8 = 2 5 ⇒ a = 2 5 , b = 4 5 .
Therefore, m i n ( a 4 + b 8 ) = 5 0 ⇒ m i n ( 1 + a 4 + b 8 ) = 1 + 5 0 = 5 1 (which I should note is greater than 3 × 5 3 4 ) :)