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

For a more geometric proof: 
Let $C$ be a cuboid whose edges are $a$, $b$ and $c$. If $\text{diam}(C)=\sqrt{a^2+b^2+c^2}$ is fixed, then $a+b+c$ is at most $\sqrt{3}\text{diam}(C)$. So $a+b+c \leq \sqrt{3(a^2+b^2+c^2)}$. 
Then, if $\text{diam}(C)$ is fixed, $\text{Vol}(C)=abc$ is at most $\displaystyle \left( \frac{a^2+b^2+c^2}{3} \right)^{3/2}$ (you maximize the volume of a cuboid inscribed into a sphere of radius $\sqrt{a^2+b^2+c^2}/2$). So $a^2+b^2+c^2 \geq 3$.
Finally, $a+b+c \leq a^2+b^2+c^2$.
