Difference between revisions of "1978 USAMO Problems/Problem 1"

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Determine the maximum value of <math>e</math>.
 
Determine the maximum value of <math>e</math>.
  
== Solution ==
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== Solution 1==
 
Accordting to '''Cauchy-Schwarz Inequalities''', we can see <math>(1+1+1+1)(a^2+b^2+c^2+d^2)\geqslant (a+b+c+d)^2</math>
 
Accordting to '''Cauchy-Schwarz Inequalities''', we can see <math>(1+1+1+1)(a^2+b^2+c^2+d^2)\geqslant (a+b+c+d)^2</math>
 
thus, <math>4(16-e^2)\geqslant (8-e)^2</math>  
 
thus, <math>4(16-e^2)\geqslant (8-e)^2</math>  
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'''from:''' [http://image.ohozaa.com/view2/vUGiXdRQdAPyw036 Image from Gon Mathcenter.net]
 
'''from:''' [http://image.ohozaa.com/view2/vUGiXdRQdAPyw036 Image from Gon Mathcenter.net]
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== Solution 2==
  
 
== See Also ==
 
== See Also ==

Revision as of 11:35, 30 April 2016

Problem

Given that $a,b,c,d,e$ are real numbers such that

$a+b+c+d+e=8$,

$a^2+b^2+c^2+d^2+e^2=16$.

Determine the maximum value of $e$.

Solution 1

Accordting to Cauchy-Schwarz Inequalities, we can see $(1+1+1+1)(a^2+b^2+c^2+d^2)\geqslant (a+b+c+d)^2$ thus, $4(16-e^2)\geqslant (8-e)^2$ Finally, $e(5e-16) \geqslant 0$ that mean, $\frac{16}{5} \geqslant e \geqslant 0$ so the maximum value of $e$ is $\frac{16}{5}$

from: Image from Gon Mathcenter.net

Solution 2

See Also

1978 USAMO (ProblemsResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5
All USAMO Problems and Solutions

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