Inequality. $a^2+b^2+c^2 \geq a+b+c$ Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that 
$a^2+b^2+c^2 \geq a+b+c$. 
Thanks

We will use the following form of Cauchy-Schwarz inequality:
From Cauchy-Schwarz inequality applied on the vectors $\displaystyle{ \left( \frac{x_1}{\sqrt{a_1}} , \frac{x_2}{\sqrt{a_2}} , \cdots , \frac{x_n}{\sqrt{a_n}} \right)}$ and $ \displaystyle{ \left( \sqrt{a_1} ,\sqrt{a_2} , \cdots , \sqrt{a_n} \right) }$ where $ x_1 ,x_2 \cdots ,x_n \in \mathbb R $ and $ a_1, a_2, \cdots ,a_n >0 $ we get that 
$ \displaystyle{ \frac{x_1^{2}}{a_1} +\frac{x_2^{2}}{a_2} + \cdots + \frac{x_n^{2}}{a_n} \geq \frac{\left(x_1 + x_2 + \cdots + x_n \right)^{2}}{a_1 +a_2 + \cdots + a_n} }$
Back to our problem now the given inequality can be written equivelant in the form 
$ \displaystyle{ \frac{a^2}{abc} + \frac{b^2}{abc} +\frac{c^2}{abc} \geq a+ b+c  \quad (\star)}$
Using now the inequality we stated in the begging we get that the left-hand-side of $(\star)$ is :
$ \displaystyle{ \frac{a^2}{abc} + \frac{b^2}{abc} +\frac{c^2}{abc} \geq \frac{ \left(a+b+c \right)^2}{3abc} = \frac{ \left(a+b+c \right)^2}{3} \geq \left(a+b+c \right) \cdot \frac{ 3 \sqrt[3]{abc}}{3} = a+b+c }$ 
which is what we need to prove. Q.E.D
P.S. I the last step we use the AM-GM inequality: $ \displaystyle{ a+b+c \geq 3 
\sqrt[3]{abc} }$.
P.S The above form of Cauchy-Schwarz inequality is called Cauchy-Schwarz in Engel form.
