Five five five and five!

Algebra Level 5

Let a , b , c , d , e R + a, b, c, d, e \in R^{+} , their sum is equal to 8 and their sum of squares is equal to 16. If the maximum value of e e can be written as A B \frac{A}{B} , when A A and B B are coprime positive integers. Find A + B A + B .


The answer is 21.

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2 solutions

Ronak Agarwal
Aug 12, 2014

We write : a + b + c + d + e = 8 a+b+c+d+e=8 and

a 2 + b 2 + c 2 + d 2 + e 2 = 16 { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 }+{ e }^{ 2 }=16 rewriting a little bit we have:

a + b + c + d = 8 e a+b+c+d=8-e (i)

a 2 + b 2 + c 2 + d 2 = 16 e 2 { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 }=16-{ e }^{ 2 } (ii)

Applying cauchy schwarz inequality we have :

4 ( a 2 + b 2 + c 2 + d 2 ) ( a + b + c + d ) 2 4({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 })\ge {(a+b+c+d)}^{2}

Substituting the values form (i) and (ii) we get :

4 ( 16 e 2 ) ( 8 e ) 2 4(16-{e}^{2})\ge{(8-e)}^{2}

Solving we get e ϵ ( 0 , 16 / 5 ] e\epsilon (0,16/5]

e m a x = 16 5 \boxed { \Rightarrow { e }_{ max }=\frac { 16 }{ 5 } }

Similar problem to mine: A difference in maxima and minima @Sanchayapol Lewgasamsarn

Sharky Kesa - 6 years, 10 months ago

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Yeah! I've seen it 3 times on the site.

Satvik Golechha - 6 years, 9 months ago

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3 times...whoa!!! So you could see the solution of one, and answer the other two :P ;)

Krishna Ar - 6 years, 9 months ago

How are you so intelligent?

Sanjana Nedunchezian - 6 years, 9 months ago

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Wow! answer @Ronak Agarwal

Satvik Golechha - 6 years, 9 months ago

You are asking me.

Ronak Agarwal - 6 years, 9 months ago
James Wilson
Dec 18, 2017

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