<s>[INST] 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?
 [/INST] Squaring the equality gives that the inequality is equivalent to
$$a^4+b^4+c^4\geq a^2+b^2+c^2.$$
This follows from Jensen applied to the convex function (defined on the positive reals) $r\mapsto r^3$, in the points $a,b,c$ and with weights $\frac{a}{a+b+c},\frac{b}{a+b+c},\frac{c}{a+b+c}$:
$$1=\left(\frac{a\cdot a+b\cdot b+c\cdot c}{a+b+c}\right)^3\leq \frac{a^3\cdot a+b^3\cdot b+c^3\cdot c}{a+b+c}$$
hence
$$a^4+b^4+c^4\geq a+b+c= a^2+b^2+c^2.$$
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