Prove this $(ab)^2+(bc)^2+(ca)^2\le ab+bc+ac$ 
Let$a$, $b$ and $c$ be non-negative numbers such that
  $$a^2+b^2+c^2=a+b+c.$$
  Show that
  $$(ab)^2+(bc)^2+(ca)^2\le ab+bc+ac.$$

I tried the uvw's technique and BW and more but without some success.I think can use C-S solve it?

Another way:
$$\sum_{cyc}(ab-a^2b^2)=\frac{1}{2}\sum_{cyc}(2ab-2a^2b^2)=$$
$$=\frac{1}{2}\sum_{cyc}(a^2+2ab-a^4-2a^2b^2+a^4-a^2)=\frac{1}{2}\sum_{cyc}(a^4-a^2)=$$
$$=\frac{1}{2}\sum_{cyc}(a^4-a^2-2(a^2-a))=\frac{1}{2}\sum_{cyc}(a^4-3a^2+2a)=\frac{1}{2}\sum_{cyc}a(a+2)(a-1)^2\geq0.$$
Done!
