$5$ real numbers tied by two equations: understanding a proof of the related problem.

Question

If $a,b,c,d,e$ are real numbers such that $$ \left\{ \begin{array}{lcl} \phantom{0}40-e &=& a+b+c+d, \\ 400-e^2 &=& a^2+b^2+c^2+d^2, \end{array} \right. $$ find $\max(e)$.

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

We know that \begin{align*} (40-e)^2 &= (a+b+c+d)^2 \\ &= a^2 + b^2 + c^2 + d^2 + 2(ab+ac+ad+bc+bd+cd). \end{align*} Now note that $2xy\leq x^2+y^2$ for all reals $x,y$. Hence $$ (40-e)^2 \leq 4(a^2+b^2+c^2+d^2). \tag{1} $$

I understand why $2xy\leq x^2+y^2$, 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.

Answer 1

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 $$2cd \leq c^2+d^2$$

Answer 2

You must write $$2ab\le a^2+b^2$$ and so on.

Answer 3

I think the following way is much more better.

By C-S $$400-e^2=a^2+b^2+c^2+d^2=\frac{1}{4}(1+1+1+1)(a^2+b^2+c^2+d^2)\geq$$ $$\geq\frac{1}{4}(a+b+c+d)^2=\frac{1}{4}(40-e)^2,$$ which gives $0\leq e\leq16.$

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