Vtyjkyuk

If $a,b,c,d,e$ are real numbers such that
$$
left{
beginarraylcl
phantom040-e &=& a+b+c+d, \
400-e^2 &=& a^2+b^2+c^2+d^2,
endarray
right.
$$
find $max(e)$.

In the solution provided, I don't understand equation $(1)$:

We know that
beginalign*
(40-e)^2
&= (a+b+c+d)^2 \
&= a^2 + b^2 + c^2 + d^2 + 2(ab+ac+ad+bc+bd+cd).
endalign*
Now note that $2xyleq x^2+y^2$ for all reals $x,y$. Hence
$$
(40-e)^2 leq 4(a^2+b^2+c^2+d^2).
tag1
$$

I understand why $2xyleq x^2+y^2$, but I'm not sure how the author derives the above inequality. $4(a^2+b^2+c^2+d^2)$ doesn't seem to represent the sum of two squares. The way the proof is worded so far implies this is obvious, so I think I'm clearly missing something. An explanation as to why the above inequality is true would be very much appreciated.

He is using it for $$2ab leq a^2+b^2$$ $$2ac leq a^2+c^2$$ $$2ad leq a^2+d^2$$ $$2bc leq b^2+c^2$$ $$2bd leq b^2+d^2$$ and $$2cd leq c^2+d^2$$

You must write $$2able a^2+b^2$$ and so on.

I think the following way is much more better.

By C-S $$400-e^2=a^2+b^2+c^2+d^2=frac14(1+1+1+1)(a^2+b^2+c^2+d^2)geq$$
$$geqfrac14(a+b+c+d)^2=frac14(40-e)^2,$$
which gives $0leq eleq16.$

The equality occurs for $(a,b,c,d,e)=(6,6,6,6,16),$ which says that $16$ is a maximal value.