<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] Here's another way.  Notice that
$$\frac{2+a}{2+b} = \frac{5a+2b+2c}{2a+5b+2c} = \frac52 -\frac32\cdot\frac{7b+2c}{2a+5b+2c} $$
Also using CS inequality ($\sum $ representing cyclic sums):
$$\sum \frac{7b+2c}{2a+5b+2c} \geqslant \frac{\left( \sum (7b+2c)\right)^2}{\sum (7b+2c)(2a+5b+2c)}= \frac{81(a+b+c)^2}{18\sum a^2+21(a+b+c)^2} \\= \frac{729}{18\sum a^2+189}$$
By the above bound, with $x = \sum a^2 \geqslant 3$, it is enough to show 
$$x \geqslant \frac{15}2- \frac32\cdot\frac{729}{18x + 189} \iff \frac{2(x+6)(x-3)}{2x+21}\geqslant 0$$
</s>