<s>[INST] Cauchy can't be applied on this inequality for this condition of a, b, c Given three numbers $a,b,c$ satisfy $0\le a,b,c\le 2$ and $a+b+c=3$.
Prove that $a^2+b^2+c^2\le \frac{1}{ab+bc+ca}+\frac{9}{2}$.
Attempt:
By trial and error, I know that the equality does not hold for $a=b=c=1$. However, the equality holds for all permutations of $(0;1;2)$. Because if the problem can be solved by Cauchy inequality, then the equality will usually (if not always) hold if $a=b=c$. My attempt is below:
This inequality is true:
$a^2+b^2+c^2\le \frac{1}{ab+bc+ca}+\frac{9}{2}$
$\Leftrightarrow \left(a+b+c\right)^2-2ab-2bc-2ca\le \frac{1}{ab+bc+ca}+\frac{9}{2}$
$\Leftrightarrow 3^2-\frac{9}{2}\le \frac{2\left(ab+bc+ca\right)^2+1}{ab+bc+ca}$ 
(after this I'll put $x=ab+bc+ca$)
$\Leftrightarrow 4.5\le \frac{2x^2+1}{x} \Leftrightarrow 2x^2+1 \ge 4.5x$
$\Leftrightarrow 2x^2-4.5x+1\ge 0$ $\Leftrightarrow x\le 0.25$ or $x\ge2$.
$\Leftrightarrow ab+bc+ca \le0.25$ or $ab+bc+ca\ge 2$.
At this point I'm stuck, is this the right way to solve?
 [/INST] Your way is right.
You need only to end it:
Let $a\geq b\geq c$.
Thus, $1\leq a\leq2$ and $$ab+ac+bc\geq a(b+c)=a(3-a)=2+(2-a)(a-1)\geq2.$$
</s>