Find the minimum value of $1+{ a }^{ 4 }+{ b }^{ 8 }$ if a and b are positive real numbers with $a{ b }^{ 2 }=5$ .

The answer is 51.

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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

Chirayu Bhardwaj
- 4 years, 11 months ago

a and b being positive integers does not satisfy the answer . They must be positive reals.

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Thanks for reporting. Fixed!

Satyajit Ghosh
- 5 years, 7 months ago

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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

Amanda Bloomer Houser
- 5 years, 7 months ago

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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

Satyajit Ghosh
- 5 years, 7 months ago

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@Satyajit Ghosh – why isnt the ans 51 at $\begin{cases} a=\sqrt{5}\\b=\sqrt[4]{5}\end{cases}$ . by simple am-gm?

Aareyan Manzoor
- 5 years, 7 months ago

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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.

Satyajit Ghosh
- 5 years, 7 months ago

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@Satyajit Ghosh – 627 new answer.

Aareyan Manzoor
- 5 years, 7 months ago

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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
- 5 years, 7 months ago

@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.

Amanda Bloomer Houser
- 5 years, 7 months ago

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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.

Satyajit Ghosh
- 5 years, 7 months ago

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There will be another question for 150. Btw people expect answer here.

Satyajit Ghosh
- 5 years, 7 months ago

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Please change the line of question to $a$ and $b$ are positive reals

Department 8
- 5 years, 7 months ago

They are not positive integers they are positive reals bangali

Kushagra Sahni
- 5 years, 7 months ago

its direct application of am gm

Dev Sharma
- 5 years, 7 months ago

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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

Satyajit Ghosh
- 5 years, 7 months ago

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kushagra and lakshya are your classmates??

Dev Sharma
- 5 years, 7 months ago

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@Dev Sharma – Kushagra is but not lakshya. But why? N try my next question

Satyajit Ghosh
- 5 years, 7 months ago

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@Satyajit Ghosh – solved......

Dev Sharma
- 5 years, 7 months ago

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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

Satyajit Ghosh
- 5 years, 7 months ago

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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, $1 + a^4 + b^8 \geq 3 \times \sqrt[3]{a^4 b^8} = 3(a b^2)^{\frac{4}{3}} = 3 \times 5^{\frac{4}{3}}$

Equality occurs when $1 = a^4 = b^8 = 5^{\frac{4}{3}}$ . But $1 = 5^{\frac{4}{3}}$ is a contradiction. So while $3 \times 5^{\frac{4}{3}}$ represents a lower bound on the expression, it does $\underline{not}$ a minimum.

Instead, apply the AM-GM Inequality on $a^4 + b^8$ .

$a^4 + b^8 \geq 2 \times \sqrt[2]{a^4 b^8}= (a b^2)^{2} = 2 \times 5^2$

Equality occurs when $a^4 = b^8 = 25 \Rightarrow a = \sqrt[2]{5}, b = \sqrt[4]{5}$ .

Therefore, $min(a^4 + b^8) = 50 \Rightarrow min(1 + a^4 + b^8) = 1+ 50 = \boxed{51}$ (which I should note is greater than $3 \times 5^{\frac{4}{3}}$ ) :)