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

Using Cauchy-Schwarz inequality we get
$$
a+b+c=a\cdot 1+b\cdot 1+c\cdot 1+\leq\sqrt{a^2+b^2+c^2}\sqrt{1^2+1^2+1^2}\tag{1}
$$
From AM-GM we obtain $a^2+b^2+c^2\geq 3\sqrt[3]{a^2b^2c^2}=3$, so
$$
\sqrt{3}\leq\sqrt{a^2+b^2+c^2}\tag{2}
$$
From $(1)$ and $(2)$ it follows
$$
a+b+c\leq\sqrt{a^2+b^2+c^2}\sqrt{3}\leq\sqrt{a^2+b^2+c^2}\sqrt{a^2+b^2+c^2}=a^2+b^2+c^2
$$
