Oh wow, this is kind of foolish problem. $$\sqrt{x + y+z+\frac{yz}{x}} +\sqrt{y + x+z+ \frac{zx}{y}} +\sqrt{z + x+y+\frac{xy}{z}} \ge\sqrt{(xy+yz+zy)(\frac{1}{x} + \frac{1}{y} + \frac{1}{z})}+\sqrt{x }+\sqrt{y} +\sqrt{z}$$$$\sqrt{\frac{(x+y)(x+z)}{x}} +\sqrt{\frac{(y+x)(y+z)}{y}} +\sqrt{\frac{(z+x)(z+y)}{z}} \ge\sqrt{(xy+yz+zy)(\frac{1}{x} + \frac{1}{y} + \frac{1}{z})}+\sqrt{x }+\sqrt{y} +\sqrt{z}$$Take $x$ as $a^2$, $y$ as $b^2$ and $z$ as $c^2$: $$\sqrt{\frac{(a^2+b^2)(a^2+c^2)}{a^2}} +\sqrt{\frac{(b^2+a^2)(b^2+c^2)}{b^2}} +\sqrt{\frac{(c^2+a^2)(c^2+b^2)}{c^2}} \ge\sqrt{((ab)^2+(bc)^2+(ca)^2)(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2})}+a+b+c$$Multiplying both sides by $abc$, we get: $$bc\sqrt{(a^2+b^2)(a^2+c^2)} +ac\sqrt{(b^2+a^2)(b^2+c^2)} +ab\sqrt{(c^2+a^2)(c^2+b^2)} \ge (ab)^2+(bc)^2+(ca)^2+a^2bc+ab^2c+abc^2,$$equivalently, $\sum_{cyc}ab\sqrt{(c^2+a^2)(c^2+b^2)} \ge \sum_{cyc}(ab)^2+abc^2$. Now, $\sqrt{(c^2+a^2)(c^2+b^2)} \ge ab+c^2\implies c^4 + (ac)^2 + (ab)^2 +(bc)^2 \ge (ab)^2 + c^4 +2abc^2\implies a^2+b^2\ge 2ab$. We are done.

parmenides51 wrote: Let $x, y, z$ be positive numbers such that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$. Prove that $$\sqrt{x + yz} +\sqrt{y + zx} +\sqrt{z + xy} \ge\sqrt{xyz}+\sqrt{x }+\sqrt{y} +\sqrt{z}$$ Given $a,b,c>0$ and $ab+bc+ca=abc$. Prove that$$\sqrt{a+bc}+\sqrt{b+ca}+\sqrt{c+ab}\ge\sqrt{abc}+\sqrt a+\sqrt b+\sqrt c.$$

parmenides51 wrote: Let $x, y, z$ be positive numbers such that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$. Prove that $$\sqrt{x + yz} +\sqrt{y + zx} +\sqrt{z + xy} \ge\sqrt{xyz}+\sqrt{x }+\sqrt{y} +\sqrt{z}$$ APMO 2002 Slovakia 2011