# real numbers

• Jul 18th 2008, 10:52 AM
perash
real numbers
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.

What is the largest possible value of e?
• Jul 19th 2008, 12:18 AM
JaneBennet
Let’s suppose $a$, $b$, $c$, $d$ are non-negative. (Obviously $e$ will have to be positive if we want to maximize it.)

Applying AM–GM to the first equation gives

$8-e=4\left(\frac{a+b+c+e}{4}\right)\ge\sqrt[4]{abcd}$

$\therefore\ e\le8-4\sqrt[4]{abcd}$

and e is maxed when $a=b=c=d$.

This is consistent with the application of AM–GM to the second equation, when we get

$e^2\le16-4\sqrt[4]{a^2b^2c^2d^2}=16-4d^2$ when $a^2=b^2=c^2=d^2$.

(Clearly, for positive e, e is maxed if and only $e^2$ is maxed.)

Solving those two equations in $e$ and $d$ gives $e=\frac{16}{5}$ as the maximum value for $a,b,c,d\ge0$.

If some or all of $a$, $b$, $c$, $d$ are negative, some other method may have to be tried.