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?

we have to show that
$$(a+b+c)^2((ab)^2+(bc)^2+(ca)^2)\le (ab+bc+ca)(a^2+b^2+c^2)^2$$
multiplying this out we get:
$${a}^{5}b+{a}^{5}c-{a}^{4}{b}^{2}+{a}^{4}bc-{a}^{4}{c}^{2}-{a}^{2}{b}^{
4}-3\,{a}^{2}{b}^{2}{c}^{2}-{a}^{2}{c}^{4}+a{b}^{5}+a{b}^{4}c+ab{c}^{4
}+a{c}^{5}+{b}^{5}c-{b}^{4}{c}^{2}-{b}^{2}{c}^{4}+b{c}^{5}
\geq 0$$
with
$$b=a+u,c=a+u+v$$ we obtain
$$\left( 9\,{u}^{2}+9\,uv+9\,{v}^{2} \right) {a}^{4}+ \left( 20\,{u}^{3
}+30\,{u}^{2}v+42\,u{v}^{2}+16\,{v}^{3} \right) {a}^{3}+ \left( 15\,{u
}^{4}+30\,{u}^{3}v+57\,{u}^{2}{v}^{2}+42\,u{v}^{3}+9\,{v}^{4} \right) 
{a}^{2}+ \left( 4\,{u}^{5}+10\,{u}^{4}v+28\,{u}^{3}{v}^{2}+32\,{u}^{2}
{v}^{3}+14\,u{v}^{4}+2\,{v}^{5} \right) a+3\,{u}^{4}{v}^{2}+6\,{u}^{3}
{v}^{3}+4\,{u}^{2}{v}^{4}+u{v}^{5}
\geq 0$$ which is true.
