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?
 A: 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
$$a^4+b^4+c^4\geq a+b+c= a^2+b^2+c^2.$$
A: 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.
A: 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!
A: 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!
