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?

I think $uvw$ helps here.
The homogenization gives:
$$(a^2b^2+a^2c^2+b^2c^2)(a+b+c)^2\leq(ab+ac+bc)(a^2+b^2+c^2)^2.$$
Hence, the condition it's $u=3u^2-2v^2$ and does not depend on $w^3$.
In another hand, we need to prove that $$3v^4-2uw^3\leq v^2,$$
which is a linear inequality of $w^3$, which says that
it remains to prove our inequality for an extremal value of $w^3$,
which happens in the following cases.


*

*$w^3=0$.


Since our new inequality is homogeneous, we can assume $c=0$ and $b=1$, which gives
$$a(a^2-a+1)(a-1)^2\geq0;$$


*$b=c=1$, which gives
$$(a-1)^2(2a^2+3a+4)\geq0.$$
Done!


But we can prove this by another way:
we need to prove that
$$\sum_{cyc}(a^4+2a^2b^2)\sum_{cyc}ab\geq\sum_{cyc}a^2b^2\sum_{cyc}(a^2+2bc)$$ or
$$\sum_{cyc}(a^5b+a^5c-a^4b^2-a^4c^2+a^4bc-a^2b^2c^2)\geq0$$ or
$$\sum_{cyc}ab(a^2-ab+b^2)(a-b)^2+\frac{1}{2}abc(a+b+c)\sum_{cyc}(a-b)^2\geq0$$
and we are done!
