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.

• what are a,b,... inetgers?natural numbers?etc. Jan 23 '15 at 10:05
• Integers, I think.But maybe it works for real too. Jan 23 '15 at 10:08

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: $$a^2+b^2+c^2+d^2+e^2\ge a(b+c+d+e)\quad\Box$$
Hint: $\frac{1}{4}a^2 + b^2 \geq ab$ and similarly for the others
Hint: complete the square for $b^2-ab$, and for $c^2-ac$.