<s>[INST] Minimum $e$ where $a,b,c,d,e$ are reals such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$ I have a question about this 1978 USAMO problem:

Given that $a,b,c,d,e$ are real numbers such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$, find the maximum value that $e$ can attain. 

I had the following solution:
Let $a+b+c+d=x$. Then $x+e=8\implies e=8-x$. Also, $\frac{a^2+1}{2}+\frac{b^2+1}{2}+\frac{c^2+1}{2}+\frac{d^2+1}{2}\geq (a+b+c+d)=x$ by AM-GM inequality. Hence, $a^2+b^2+c^2+d^2\geq 2x-4$. 
Now we have $2x-4+e^2\leq a^2+b^2+c^2+d^2+e^2=16$. Substituting $x=8-e$, we get $e^2-2e-4\leq 0$. We can easily calculate that the lowest value that $e$ can attain is $1 -\sqrt{5}$. 
However, the answer given on the internet is $\frac{16}{5}$. Where am I going wrong?
EDIT $1$ -- Is this a case of how the value $1-\sqrt{5}$ can never be attained by $e$, although the inequality is true?
EDIT $2$ -- It seems that we need to find the maximum. By my method, I’ve found the maximum to be $1+\sqrt{5}$. This is greater than $\frac{16}{5}$. Have I found a sharper inequality? 
 [/INST] Given $a+b+c+d+e=8$ and $a+b+c+d=8-e$
$$(8-e)^2=(a+b+c+d)^2$$
Now expand $(a+b+c+d)^2$
$$=a^2+b^2+c^2+d^2+2(ab+ac+ad+bc+bd+cd)$$
$$\le4(a^2+b^2+c^2+d^2)$$
And Now, $a^2+b^2+c^2+d^2+e^2=16$ 
$a^2+b^2+c^2+d^2=16-e^2$ $$(16-e^2)=a^2+b^2+c^2+d^2$$$$4(a^2+b^2+c^2+d^2)\ge(a+b+c+d)^2$$$$4(16-e^2)\ge(8-e)^2$$$$64-4e^2\ge64-16e^2+e^2$$$$5e^2-16e\le0$$$$e(5e-16)\le0$$$$e\le\dfrac{16}{5}$$
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