<s>[INST] Prove : $\sqrt{\frac{a^2+b^2+1}{a^2+c^2+1}}+\sqrt{\frac{a^2+c^2+1}{b^2+c^2+1}}+\sqrt{\frac{b^2+c^2+1}{a^2+b^2+1}}\geq 3$ with $a+b+c=1$ Claim :


Let $a\geq b\geq c\geq 0$ and $a+b+c=1$ then we have :
$$\sqrt{\frac{a^2+b^2+1}{a^2+c^2+1}}+\sqrt{\frac{a^2+c^2+1}{b^2+c^2+1}}+\sqrt{\frac{b^2+c^2+1}{a^2+b^2+1}}\geq 3$$


To prove it I try the Ji chen's theorem because we have :
$$\frac{a^2+b^2+1}{a^2+c^2+1}+\frac{a^2+c^2+1}{b^2+c^2+1}+\frac{b^2+c^2+1}{a^2+b^2+1}\geq 3$$
$$\frac{a^2+b^2+1}{a^2+c^2+1}\frac{a^2+c^2+1}{b^2+c^2+1}+\frac{a^2+b^2+1}{a^2+c^2+1}\frac{b^2+c^2+1}{a^2+b^2+1}+\frac{a^2+c^2+1}{b^2+c^2+1}\frac{b^2+c^2+1}{a^2+b^2+1}\geq 3$$
$$\frac{a^2+b^2+1}{a^2+c^2+1}\frac{a^2+c^2+1}{b^2+c^2+1}\frac{b^2+c^2+1}{a^2+b^2+1}=1$$
To prove the two first I use Karamata's inequality as we have .


Let $x,y,z,u,v,w>0$ such that $x\geq y \geq z$ and $u\geq v \geq w$ and :
$x\geq u$,
$xy\geq uv$,
$xyz\geq uvw$
Then we have :
$$x+y+z\geq u+v+w$$


Question :
Is there simpler ?
Thanks in advance !
Regards
Max
 [/INST] Hint:clearly this is of form $$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\ge 3$$ which is true by am-gm
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