<s>[INST] If $a+b+c=3$ Prove that $a^{2}+b^{2}+c^{2}\geq\frac{2+a}{2+b}+\frac{2+b}{2+c}+\frac{2+c}{2+a}$ Question -

Let $a, b, c$ be positive real numbers such that $a+b+c=3 .$ Prove that
  $$
a^{2}+b^{2}+c^{2} \geq \frac{2+a}{2+b}+\frac{2+b}{2+c}+\frac{2+c}{2+a}
$$

My try -
i tried putting $a+2 = x, b+2=y , c+2=z$
then we get $x+y+z=9$ and after simplification we have to prove that 
$3>x/y + y/z + z/x$ which i am not able to prove...
i also tried C-S,Chebyshev,rearrangement..etc but none of them working
any hints ???
thankyou
 [/INST] Let $c=\min\{a,b,c\}$.
Thus, we need to prove that:
$$a^2+b^2+c^2-\frac{(a+b+c)^2}{3}\geq\frac{2+a}{2+b}+\frac{2+b}{2+a}-2+\frac{2+b}{2+c}-\frac{2+b}{2+a}+\frac{2+c}{2+a}-1$$ or
$$\frac{2}{3}((a-b)^2+(c-a)(c-b))\geq\frac{(a-b)^2}{(2+a)(2+b)}+\frac{(c-a)(c-b)}{(2+a)(2+c)},$$ which is true because
$$\frac{2}{3}>\frac{1}{4}>\frac{1}{(2+a)(2+b)}$$ and
$$\frac{2}{3}>\frac{1}{(2+a)(2+c)}.$$
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