Pinko wrote: The real numbers $a$, $b$, $c$ are such that $a+b+c+ab+bc+ca+abc \geq 7$. Prove that $\sqrt{a^2+b^2+2}+\sqrt{b^2+c^2+2}+\sqrt{c^2+a^2+2} \geq 6$ Very nice! Let us rewrite the condition in the following form: $(a+1)(b+1)(c+1)\geq 8$. Put $x=a^2+1$, $y=b^2+1$ and $z=c^2+1$, then $$xyz=(a^2+1)(b^2+1)(c^2+1)\geq \frac{\Big((a+1)(b+1)(c+1)\Big)^2}{8}=8.$$Now we use the Cauchy inequality: $$\sqrt{a^2+b^2+2}+\sqrt{b^2+c^2+2}+\sqrt{c^2+a^2+2}=\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\geq 3\sqrt[6] {(x+y)(y+z)(z+x)}\geq 3\sqrt[6]{8xyz}=6.$$

I think this is same as A1 JBMO Shortlist 2017. Rewrite the condition as $(a + 1)(b + 1)(c + 1) \ge 8$, and using the fact that \[ a^2 + b^2 + 2 \ge ab + a + b + 1 = (a + 1)(b + 1) \]We'll get \[ \sqrt{a^2 + b^2 + 2} + \sqrt{b^2 + c^2 + 2} + \sqrt{c^2 + a^2 + 2} \ge \sqrt{(a + 1)(b + 1)} + \sqrt{(b + 1)(c + 1)} + \sqrt{(c + 1)(a + 1)} \ge 3 \sqrt[3]{(a + 1)(b + 1)(c + 1)} \ge 6 \]

Pinko wrote: The real numbers $a$, $b$, $c$ are such that $a+b+c+ab+bc+ca+abc \geq 7$. Prove that $\sqrt{a^2+b^2+2}+\sqrt{b^2+c^2+2}+\sqrt{c^2+a^2+2} \geq 6$ Proof of Zhangyunhua:

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