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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Tags: AMC, USA(J)MO, USAMO, inequalities, vector
16.08.2011 22:03
18.08.2011 07:35
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?
18.08.2011 08:30
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.
20.07.2012 16:32
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 $a_5$.
20.07.2012 16:59
$a+b+c+d=8-e$ 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 $4(16-e^2)\ge(8-e)^2$
20.07.2012 17:48
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).
21.07.2012 08:29
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.
21.07.2012 10:06
, 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.
07.11.2013 02:48
bzprules wrote:
What is the motivation for this? How does one think to use C-S on just a, b, c, d? Thanks
12.11.2013 03:27
\bump...Sorry for bumping, but I want to learn how to better see what to do on inequality problems Thanks
12.11.2013 16:14
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!
13.11.2013 04:35
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.
20.11.2013 08:33
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 $a, b, c, d$.
27.11.2013 02:33
This is a solution I saw in a Inequalities handout somewhere (I think Kedlaya) and thought it would be interesting to put it here.
26.04.2014 03:26
$e^2=16-(a^2+b^2+c^2+d^2)\le 16-\frac{1}{4}(|a|+|b|+|c|+|d|)^2$ $\le 16-\frac{1}{4}|a+b+c+d|^2=4e-\frac{1}{4}e^2\Rightarrow$ $0 \le e \le \frac{16}{5}$. See also here http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=353574&hilit=inequality
18.01.2016 02:55
MathNewb wrote: Is there any proof available (or the name of the inequality that I could look up?) that in general: $$ \sum_{i=1}^n x_i^2 \ge \frac{1}{n}\left(\sum_{i=1}^n x_i\right)^2 $$ http://www.artofproblemsolving.com/community/c6h1186121p5765803
23.03.2021 11:46
First Solution : We know $a+b+c+d+e=8\Longleftrightarrow a+b+c+d=8-e$ and $a^2+b^2+c^2+d^2+e^2=16\Longleftrightarrow a^2+b^2+c^2+d^2=16-e^2$ (*). Next we know $\left(\dfrac{a+b+c+d}{4}\right)^2\le\dfrac{a^2+b^2+c^2+d^2}{4}$ using Tchebycheff's Inequality. Last, $$\begin{array}{lll} \left(\dfrac{8-e}{4}\right)^2&\le&\dfrac{16-e^2}{4}\\ 64+e^2-16e&\le&4(16-e^2)\\ e(5e-16)&\le&0\\ e&\le&\dfrac{16}{5} \end{array}$$Thus maximum of $e$ is $\dfrac{16}{5}$ Second Solution : With arithmetic mean and quadratic mean : $$\dfrac{a+b+c+d}{4}\le\sqrt{\dfrac{a^2+b^2+c^2+d^2}{4}}$$From (*) in First Solution, we have : $$\begin{array}{lll} \dfrac{8-e}{4}&\le&\sqrt{\dfrac{16-e^2}{4}}\\ 64+e^2-16e&\le&4(16-e^2)\\ e(5e-16)&\le&0\\ e&\le&\dfrac{16}{5} \end{array}$$Thus maximum of $e$ is $\dfrac{16}{5}$ Rafa171206
23.03.2021 13:14
More generally $a_1+a_2+...+a_n=x>0 , a_1^2+a_2^2+...+a_n^2=y>0$ Maximum value of $a_n$ ?
23.03.2021 13:31
Moubinool wrote: More generally $a_1+a_2+...+a_n=x>0 , a_1^2+a_2^2+...+a_n^2=y>0$ Maximum value of $a_n$ ? Use C-S.
24.10.2023 20:47
We have $a+b+c+d=8-e$ and $a^2+b^2+c^2+d^2=16-e^2$. Thus, \[(1+1+1+1)(a^2+b^2+c^2+d^2) \ge (a+b+c+d)^2 \implies 4(16-e^2) \ge (8-e)^2\] which gives us $e \le \boxed{\frac{16}{5}}$, which is achievable.
07.11.2023 01:45
26.11.2023 06:16
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 $0 \le e \le \boxed{\tfrac{16}5}$.
21.01.2024 18:23
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
21.02.2024 07:37
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 $\boxed{\frac{16}{5}}$. $\blacksquare$
21.02.2024 07:43
the amount of identical solutions posted in this thread is crazy
21.02.2024 08:12
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 $f\left(\frac{6}{5}\right)=\frac{16}{5}$.
21.02.2024 08:14
For the sake of an original solution, we won't use Cauchy. Instead, we utilize the ingenious Titu's Inequality \[16-e^2 = \frac{a^2}{1} + \frac{b^2}{1} + \frac{c^2}{1} + \frac{d^2}{1} \ge \frac{(a+b+c+d)^2}{4} = \frac{(8-e)^2}{4}\]\[\implies 5e^2 - 16e \leq 0 \implies e \in \left[0, \frac{16}{5}\right],\] where the equality case is indeed attainable when $a=b=c=d=\tfrac 65$. Hence our answer is $\boxed{\frac{16}{5}}$. $\blacksquare$
10.03.2024 01:21
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.
12.06.2024 04:21
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.
12.06.2024 04:55
torch wrote: the amount of identical solutions posted in this thread is crazy
but whatever at least it's original
22.08.2024 21:57
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 $\boxed{\frac{16}{5}}$.
08.09.2024 04:40
By AM-QM inequality we have $$\frac{8-e}{4} \leq \sqrt{\frac{16-e^2}{4}}$$or $e \leq \boxed{\frac{16}{5}}$
03.11.2024 17:13
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 $\boxed{e = \frac{16}{5}}$.
02.01.2025 00:48
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