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Mrdavid445

5123 posts

Mrdavid445

#1 h Aug 16, 2011, 6:32 PM • 5 Y

Y by samrocksnature, Adventure10, Mango247, kiyoras_2001, NicoN9

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 $e$ .

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GlassBead

1583 posts

GlassBead

#2 h Aug 16, 2011, 7:03 PM • 13 Y

Y by BakyX, SBS1, andrei_cdr29, samrocksnature, ehuseyinyigit, zhoujef000, Adventure10, Mango247, kiyoras_2001, NicoN9, and 3 other users

Solution

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JwooSon

8 posts

JwooSon

#3 h Aug 18, 2011, 4:35 AM • 3 Y

Y by samrocksnature, Adventure10, Mango247

Does this mean that whatever's inside the sigma doesn't necessarily have to be a sequence, since we don't know if a, b, c, and d are in any arithmetic or geometric sequence?

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fishythefish

4001 posts

fishythefish

#4 h Aug 18, 2011, 5:30 AM • 3 Y

Y by JwooSon, samrocksnature, Adventure10

I'm assuming you're asking about Cauchy-Schwarz? Then, yes, the summands in the C-S inequality are arbitrary numbers (actually, vectors in an inner product space - but these are just numbers for the reals), not necessarily terms of a sequence.

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SBS1

19 posts

SBS1

#5 h Jul 20, 2012, 1:32 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

Equivalently ,Given that $a_i,b_i,c_i,d_i,e_i$ are real numbers such that

$\sum_{i=1}^5 a_i=8$ , and $\sum _{i=1}^5 a_i^2=16$ .

Determine the maximum value of $a_5$ .

Another Solution

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FrankScience

8 posts

FrankScience

#6 h Jul 20, 2012, 1:59 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

$a+b+c+d=8-e$ and $a^2+b^2+c^2+d^2=16-e^2$
$4(a^2+b^2+c^2+d^2)\ge(a+b+c+d)^2$ (Cauchy-Schwarz Inequality)
Therefore
$4(16-e^2)\ge(8-e)^2$

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SBS1

19 posts

SBS1

#7 h Jul 20, 2012, 2:48 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

FrankScience wrote:

$a+b+c+d=8-e$ and $a^2+b^2+c^2+d^2=16-e^2$
$4(a^2+b^2+c^2+d^2)\ge(a+b+c+d)^2$ (Cauchy-Schwarz Inequality)
Therefore
$4(16-e^2)\ge(8-e)^2$

That's exactly what bzprules did (above).

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mathbuzz

803 posts

mathbuzz

#8 h Jul 21, 2012, 5:29 AM • 3 Y

Y by samrocksnature, Adventure10, Mango247

apply that by CS inequality , $4(a^2+b^2+c^2+d^2)$ $\ge$ $(a+b+c+d)^2$
so , $4(16-e^2)$ $\ge$ $(8-e)^2$ . then it is trivial.

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mavropnevma

15142 posts

mavropnevma

#9 h Jul 21, 2012, 7:06 AM • 15 Y

Y by AwesomeToad, Binomial-theorem, bluecarneal, dantx5, joybangla, fractals, eshan, raxu, volcanogobbler, OlympusHero, samrocksnature, Adventure10, Mango247, aidan0626, balllightning37

A festival of copycats? Or maybe just carelessness? Although bzprules (the first who posted a valid solution) hid his under the name Solution, FrankScience, and later mathbuzz, posted sketches on precisely the same idea. Now, it may be that the first did not bother to read what was written under the hiding tag, but now his half-solution was in plain view, so the second does not even have this excuse.

In general, it is recommended, when contributig to a topic, to check if what you have to say has not already been said. Now, even if the general idea is the same, bringing new sides to the story, turning an unturned-yet little pebble, shedding some more light on a more obscure corner, making a comment, linking to another problem on same grounds, is useful and welcomed. But bluntly just spitting the same formulae is useless, and frown upon.

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supercomputer

491 posts

supercomputer

#10 h Nov 7, 2013, 12:48 AM • 3 Y

Y by samrocksnature, Adventure10, Mango247

bzprules wrote:

Solution

What is the motivation for this? How does one think to use C-S on just a, b, c, d?

Thanks

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supercomputer

491 posts

supercomputer

#11 h Nov 12, 2013, 1:27 AM • 2 Y

Y by samrocksnature, Adventure10

\bump...Sorry for bumping, but I want to learn how to better see what to do on inequality problems
Thanks

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arqady

30150 posts

arqady

#12 h Nov 12, 2013, 2:14 PM • 3 Y

Y by samrocksnature, Adventure10, and 1 other user

supercomputer wrote:

I want to learn how to better see what to do on inequality problems
Thanks

There are about 30 methods for proving inequalities. Learn them!

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supercomputer

491 posts

supercomputer

#13 h Nov 13, 2013, 2:35 AM • 2 Y

Y by samrocksnature, Adventure10

No...I know the methods; I am wondering how someone thought to consider C-S on $a,b,c,d$ (as opposed to some other group of variables) for this specific problem.

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fractals

3028 posts

fractals

#14 h Nov 20, 2013, 6:33 AM • 5 Y

Y by supercomputer, samrocksnature, Adventure10, Mango247, and 1 other user

It's because you want $e$ isolated; you want that variable maximized, which will be at some extremum probably not symmetric with the rest (so the standard stuff shouldn't work). But probably $a, b, c, d$ will be equal to "maximize" it. So apply it on $a, b, c, d$ .

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ssilwa

5451 posts

ssilwa

#15 h Nov 27, 2013, 12:33 AM • 4 Y

Y by biomathematics, samrocksnature, Adventure10, Mango247

This is a solution I saw in a Inequalities handout somewhere (I think Kedlaya) and thought it would be interesting to put it here.

Click to reveal hidden text

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zhoujef000

285 posts

zhoujef000

#41 h Nov 6, 2023, 10:45 PM

Y by

completing the square???

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pqr.

174 posts

pqr.

#42 h Nov 26, 2023, 3:16 AM • 1 Y

Y by MihaiT

Weird problem.

Note that $e=8-(a+b+c+d)$ . By Cauchy,
$(a^2+b^2+c^2+d^2)(1^2+1^2+1^2+1^2) \ge (a+b+c+d)^2$ $\implies 4(16-e^2) \ge (8-e)^2,$ and solving gives $0 \le e \le \boxed{\tfrac{16}5}$ .

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bjump

965 posts

bjump

#43 h Jan 21, 2024, 3:23 PM • 1 Y

Y by MihaiT

We claim the maximum $e$ is $\frac{16}{5}$ , which can be obtained by $(a,b,c,d,e)= (\tfrac{6}{5},\tfrac{6}{5},\tfrac{6}{5},\tfrac{6}{5}, \tfrac{16}{5})$
Note that
$a+b+c+d=8-e$ $a^{2}+b^{2}+c^{2}+d^{2}=16-e^{2}$ By Cauchy-Schwarz
$(a^{2}+b^{2}+c^{2}+d^{2})(1+1+1+1) \geq (a+b+c+d)^{2}$ $4(16-e^{2}) \geq (8-e)^{2}$ $0\geq 5e^{2}-16e$ $\frac{16}{5} \geq e \geq 0$ Finishing the proof

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Math4Life7

1703 posts

Math4Life7

#44 h Feb 21, 2024, 4:37 AM

Y by

We get from cauchy that $(4)(a^2+b^2+c^2+d^2) \geq (a+b+c+d)^2$ . This means that $4(16-e^2) \geq 8 - e$ . This gives $5e^2 - 16e \geq 0$ . THis obviously gives maximal value of $\boxed{\frac{16}{5}}$ . $\blacksquare$

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torch

996 posts

torch

#45 h Feb 21, 2024, 4:43 AM

Y by

the amount of identical solutions posted in this thread is crazy

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anduran

463 posts

anduran

#46 h Feb 21, 2024, 5:12 AM

Y by

solution by Thomas Mildorf:
Observe that the givens can be effectively combined by considering squares:
$(a-k)^2+(b-k)^2+(c-k)^2+(d-k)^2+(e-k)^2=(a^2+b^2+c^2+d^2+e^2)-2r(a+b+c+d+e)+5r^2=16-16r+5r^2$ Since squares are nonnegative, we have
$e \leq \sqrt{5r^2-16r+16}+r=f(r)$ And equality can occur at $a=b=c=d=r$ , so all that remains is to compute the minimum of $f(r)$ .
Since $f(r)$ grows large at both $r=\pm \infty$ , the minimum occurs at $f'(r)=1+\frac{10r-16}{2\sqrt{5r^2-16r+16}}=0$ . Solving yields $r=\frac{6}{5}, 2$ , but $2$ is an extraneous solution, hence the minimum of $f(r)$ , or the maximum of $e$ , occurs at $f\left(\frac{6}{5}\right)=\frac{16}{5}$ .

This post has been edited 1 time. Last edited by anduran, Feb 21, 2024, 5:12 AM

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shendrew7

792 posts

shendrew7

#47 h Feb 21, 2024, 5:14 AM

Y by

For the sake of an original solution, we won't use Cauchy. Instead, we utilize the ingenious Titu's Inequality :coolspeak:

$16-e^2 = \frac{a^2}{1} + \frac{b^2}{1} + \frac{c^2}{1} + \frac{d^2}{1} \ge \frac{(a+b+c+d)^2}{4} = \frac{(8-e)^2}{4}$ $\implies 5e^2 - 16e \leq 0 \implies e \in \left[0, \frac{16}{5}\right],$
where the equality case is indeed attainable when $a=b=c=d=\tfrac 65$ . Hence our answer is $\boxed{\frac{16}{5}}$ . $\blacksquare$

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RedFireTruck

4216 posts

RedFireTruck

#48 h Mar 9, 2024, 10:21 PM

Y by

trivial by Jensen's

We want to maximize $e$ so that there exists a solution to $a+b+c+d=8-e$ and $a^2+b^2+c^2+d^2=16-e^2$ . Since $y=x^2$ is convex, we use Jensen's Inequality to get that the minimum value of $a^2+b^2+c^2+d^2$ when $a+b+c+d=8-e$ is $4(2-\frac{e}4)^2=\frac{e^2}{4}-4e+16$ with equality at $a=b=c=d=2-\frac{e}4$ . Therefore, we want the maximum $e$ such that $\frac{e^2}{4}-4e+16\le 16-e^2$ or $5e^2\le 16e$ . We see that the maximum possible value of $e$ is $\boxed{\frac{16}{5}}$ with $a=b=c=d=\frac65$ as a solution.

This post has been edited 1 time. Last edited by RedFireTruck, Mar 9, 2024, 10:23 PM

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blueprimes

303 posts

blueprimes

#49 h Jun 12, 2024, 1:21 AM

Y by

A classic one.

We claim the maximum is $\frac{16}{5}$ , attainable at $(a, b, c, d, e) = \left(\frac{6}{5}, \frac{6}{5}, \frac{6}{5}, \frac{6}{5}, \frac{16}{5} \right)$ . To prove this is precisely the maximum, we rewrite the equations as $a + b + c + d = 8 - e$ , $a^2 + b^2 + c^2 + d^2 = 16 - e^2$ . By Cauchy-Schwarz Inequality, we obtain $4(16 - e^2) \le (8 - e)^2 \iff e \le \frac{16}{5}$ . We are done.

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DouDragon

1728 posts

DouDragon

#50 h Jun 12, 2024, 1:55 AM • 1 Y

Y by ehuseyinyigit

torch wrote:

the amount of identical solutions posted in this thread is crazy

breaking the pattern
ngl kind of a fakesolve since i don't know too much about spoiler but whatever at least it's original

This post has been edited 1 time. Last edited by DouDragon, Jun 12, 2024, 2:50 AM
Reason: latex fix

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franzliszt

23531 posts

franzliszt

#51 h Aug 22, 2024, 6:57 PM • 1 Y

Y by DouDragon

Here is a solution using Lagrange multipliers.

Given the constraints
$g(a, b, c, d, e) = a + b + c + d + e - 8 = 0$ $h(a, b, c, d, e) = a^2 + b^2 + c^2 + d^2 + e^2 - 16 = 0$ we want to maximize the objective function $f(a, b, c, d, e) = e.$
Introduce multipliers $\lambda$ and $\mu$ so that $\nabla f = \lambda \nabla g + \mu \nabla h.$
The gradients are
$\begin{align*} \nabla f &= (0, 0, 0, 0, 1)\\ \nabla g &= (1, 1, 1, 1, 1)\\ \nabla h &= (2a, 2b, 2c, 2d, 2e) \end{align*}$
so we have $(0, 0, 0, 0, 1) = \lambda (1, 1, 1, 1, 1) + \mu (2a, 2b, 2c, 2d, 2e).$
This gives us the system
$\begin{align*} \lambda + 2\mu a &= 0 \\ \lambda + 2\mu b &= 0 \\ \lambda + 2\mu c &= 0 \\ \lambda + 2\mu d &= 0 \\ \lambda + 2\mu e &= 1 \\ \end{align*}$
From the first four equations, we know $\lambda = -2\mu a = -2\mu b = -2\mu c = -2\mu d$ so $a=b=c=d$ . Let $a = b = c = d = x$ for some $x$ . Substituting back into the constraints gives
$\begin{align*} 4x + e &= 8\\ 4x^2 + e^2 &= 16 \end{align*}$ Solving the system for $x$ gives critical points of $x = 2$ and $x = \frac{6}{5}$ .

For $x=2$ we have $e = 8 - 4(2) = 0$ , and for $x = \frac{6}{5}$ we have $e = 8 - 4\left(\frac{6}{5}\right) = 8 - \frac{24}{5} = \frac{40}{5} - \frac{24}{5} = \frac{16}{5}$ .

Therefore, the maximum value of $e$ is $\boxed{\frac{16}{5}}$ .

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eg4334

602 posts

eg4334

#52 h Sep 8, 2024, 1:40 AM

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By AM-QM inequality we have $\frac{8-e}{4} \leq \sqrt{\frac{16-e^2}{4}}$ or $e \leq \boxed{\frac{16}{5}}$

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megahertz13

3168 posts

megahertz13

#53 h Nov 3, 2024, 2:13 PM

Y by

Note that $a+b+c+d=8-e$ and $a^2+b^2+c^2+d^2=16-e^2.$ Squaring the first equation gives $a^2+b^2+c^2+d^2+2(ab+ac+\dots+cd)=64-16e+e^2.$ Since $a^2+b^2+c^2+d^2=16-e^2$ , we know that $16-e^2+2(ab+ac+\dots+cd)=64-16e+e^2,$ or $8e+ab+ac+\dots+cd=24+e^2.$ Note that $ab+ac+\dots+cd\le \frac{3}{2}(a^2+b^2+c^2+d^2) = 24-\frac{3}{2}e^2.$ To maximize $e$ , we need to maximize $ab+ac+\dots+cd$ , so we set it to $24-\frac{3}{2}e^2$ . Solving the resulting equation gives $\boxed{e = \frac{16}{5}}$ .

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P162008

94 posts

P162008

#54 h Jan 1, 2025, 9:48 PM

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Since a,b,c,d and e are real numbers such that a + b + c + d + e = 8 and a² + b² + c² + d² + e² = 16.

Then, by Cauchy Schwarz Inequality we've

(a² + b² + c² + d²)(1 + 1 + 1 + 1) ≥ (a + b + c + d)²

4(16 - e²) ≥ (8 - e)²

5e² - 16e ≥ 0

e(5e - 16) ≥ 0

So, 16/5 ≥ e ≥ 0 which implies that the maximum value of e = 16/6

This post has been edited 1 time. Last edited by P162008, Jan 1, 2025, 9:48 PM
Reason: Typo

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Marcus_Zhang

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Marcus_Zhang

#55 h Yesterday at 6:37 PM

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Separate e