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?

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.

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 $a_5$.

$a+b+c+d=8-e$ 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 $4(16-e^2)\ge(8-e)^2$

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).

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.

, 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.

bzprules wrote:

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

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

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!

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.

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 $a, b, c, d$.

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

$e^2=16-(a^2+b^2+c^2+d^2)\le 16-\frac{1}{4}(|a|+|b|+|c|+|d|)^2$ $\le 16-\frac{1}{4}|a+b+c+d|^2=4e-\frac{1}{4}e^2\Rightarrow$ $0 \le e \le \frac{16}{5}$. See also here http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=353574&hilit=inequality

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

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

More generally $a_1+a_2+...+a_n=x>0 , a_1^2+a_2^2+...+a_n^2=y>0$ Maximum value of $a_n$ ?

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.

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.

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 $0 \le e \le \boxed{\tfrac{16}5}$.

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

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 $\boxed{\frac{16}{5}}$. $\blacksquare$

the amount of identical solutions posted in this thread is crazy

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 $f\left(\frac{6}{5}\right)=\frac{16}{5}$.

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 $\boxed{\frac{16}{5}}$. $\blacksquare$

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.

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.

torch wrote: the amount of identical solutions posted in this thread is crazy

but whatever at least it's original

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 $\boxed{\frac{16}{5}}$.

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

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 $\boxed{e = \frac{16}{5}}$.

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