A question of range.

Question

If $a,b,c,d,e$ are positive reals, such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$,then the range of $e$ is?

I dont even know how to proceed with problems like these. I have an idea that $A.M-G.M$ might be helpful. Is it true? If yes, then how to go about it?

Another one: https://math.stackexchange.com/questions/1789371/max-of-e-with-abcde-8-and-a2b2c2d2e2-16. – Both found instantly with Approach0 — Martin R, Nov 29 '17 at 15:03

nonuser · Accepted Answer · 2017-11-29T14:22:35.367

2

Use Cauchy inequality:

$$ 16-e^2 = a^2+b^2+c^2+d^2\geq {1\over 4}(a+b+c+d)^2 ={1\over 4} (8-e)^2$$

So we have $$5e^2-16e\leq 0\Longrightarrow e\in (0,{16\over 5}] $$

edited Nov 29 '17 at 14:22

answered Nov 29 '17 at 14:03

nonuser

90,026

what exactly is the cauchy inequality? any other methods other than cauchy inequality? – Yami Kanashi Nov 29 '17 at 14:15
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Quadratic mean - arithmetic mean – nonuser Nov 29 '17 at 14:16
what is a quadratic mean? – Yami Kanashi Nov 29 '17 at 14:17
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https://en.wiktionary.org/wiki/quadratic_mean – nonuser Nov 29 '17 at 14:18
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$$Q.M. \geq A.M. $$ – nonuser Nov 29 '17 at 14:19
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@JohnWatson: I think zero should be excluded – Vasili Nov 29 '17 at 14:22
@Vasya why should zero be excluded? – Yami Kanashi Nov 29 '17 at 15:33
@LokeshSangewar: $e$ is supposed to be a positive real, right? – Vasili Nov 29 '17 at 15:53
@Vasya ok so since zero isnt a positive real it should be excluded right? – Yami Kanashi Nov 29 '17 at 16:02
@LokeshSangewar: that's right – Vasili Nov 29 '17 at 16:32

A question of range.

1 Answers1