Prove that $a^2+b^2+c^2+d^2+e^2 > a(b+c+d+e)$ Prove that $a^2+b^2+c^2+d^2+e^2 > a(b+c+d+e)$
Seems to be easy but, cannot see the method right now.
Tried adding known things like $a^2+b^2>=2ab$ and so on with other letters.Maybe I didn't found the right combination.
Please give me only a little hint.

First, notice that
$$\left(\dfrac{a}{2}-b\right)^2\ge 0\Longleftrightarrow \dfrac{a^2}{4}+b^2\ge ab$$
Apply the same property to the other variables:
$$\dfrac{a^2}{4}+c^2\ge ac$$
$$\dfrac{a^2}{4}+d^2\ge ad$$
$$\dfrac{a^2}{4}+e^2\ge ae$$
Summing up everything:
$$a^2+b^2+c^2+d^2+e^2\ge a(b+c+d+e)\quad\Box$$
