Should be $x+y+z $

On the First Suggested Version: Let $xy=c$, $yz=a$ and $zx=b$ The condition becomes: $$a+b+c=ab+bc+ca$$And the inequality is equivalent to: $$ab+bc+ca\geqslant \sum \sqrt {\frac{x^3y^3z^2+x^2y^2z^2}{2}}=\sum \sqrt {\frac{a^2bc+abc}{2}}$$Now by CS $\sqrt{2}RHS=\sum \sqrt{a^2bc+abc}=\sqrt{ab(ac+c)}+\sqrt{bc(ba+a)}+\sqrt{ca(cb+b)}\leqslant \\ \leqslant\sqrt{(ab+bc+ca)(ac+ba+cb+a+b+c)}=\sqrt{2}LHS$

So from AM-GM inequality we get: $$\frac{(x+\frac{1}{y})}{2}+y \geq 2\sqrt{(x+\frac{1}{y}). \frac{y}{2}}=2\sqrt{\frac{xy+1}{2}}...(1)$$$$\frac{(y+\frac{1}{z})}{2}+z \geq 2\sqrt{(y+\frac{1}{z}). \frac{z}{2}}=2\sqrt{\frac{yz+1}{2}}...(2)$$$$\frac{(z+\frac{1}{x})}{2}+x \geq 2\sqrt{(z+\frac{1}{x}). \frac{x}{2}}=2\sqrt{\frac{zx+1}{2}}...(3)$$$\Longrightarrow$ by $(1)+(2)+(3)$ we get: \[x+y+z\geq \sqrt{\frac{xy+1}{2}}+\sqrt{\frac{yz+1}{2}}+\sqrt{\frac{zx+1}{2}} \ .\]

On the Second Suggested Version: $x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\implies \sum \frac{x^2+1}{x}=2(x+y+z)$ Now $\sum \sqrt{ \frac{x^2+1}{2}}=\sum \sqrt{\frac{x^2+1}{x}\cdot \frac{x}{2}}\leqslant \sqrt {(2\sum x)(\frac{1}{2}\sum x)}=x+y+z$

WRONG SOLUTION

@above $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1\cdot yz}{x\cdot yz}+\frac{1\cdot xz}{y\cdot xz}+\frac{1\cdot xy}{z\cdot xy}=\frac{xy+yz+zx}{xyz}$ The rest of your proof is flawed too

Where can we find the shortlist?

this was proposed by Azerbaijan

IstekOlympiadTeam wrote: Let $x,y,z$ be positive real numbers such that $x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}.$ Prove that \[x+y+z\geq \sqrt{\frac{xy+1}{2}}+\sqrt{\frac{yz+1}{2}}+\sqrt{\frac{zx+1}{2}} \ .\]

Proof of Zhangyunhua:

FabrizioFelen wrote: So from AM-GM inequality we get: $$\frac{(x+\frac{1}{y})}{2}+y \geq 2\sqrt{(x+\frac{1}{y}). \frac{y}{2}}=2\sqrt{\frac{xy+1}{2}}...(1)$$$$\frac{(y+\frac{1}{z})}{2}+z \geq 2\sqrt{(y+\frac{1}{z}). \frac{z}{2}}=2\sqrt{\frac{yz+1}{2}}...(2)$$$$\frac{(z+\frac{1}{x})}{2}+x \geq 2\sqrt{(z+\frac{1}{x}). \frac{x}{2}}=2\sqrt{\frac{zx+1}{2}}...(3)$$$\Longrightarrow$ by $(1)+(2)+(3)$ we get: \[x+y+z\geq \sqrt{\frac{xy+1}{2}}+\sqrt{\frac{yz+1}{2}}+\sqrt{\frac{zx+1}{2}} \ .\] Nice

I have a different solution: From Andreescu $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}$ thus, $(x+y+z)^2\geq 9$ We just need to prove that $(x+y+z)^2\geq (\sqrt{\frac{xy+1}{2}}+\sqrt{\frac{yz+1}{2}}+\sqrt{\frac{zx+1}{2}})^2$ From C-S we have $RHS\leq (xy+1+yz+1+zx+1)(\frac{1}{2}+\frac{1}{2}+\frac{1}{2})=\frac{3}{2}(xy+yz+zx+3)$ So the given inequality is equivalent to: $2(x+y+z)^2\geq 3(xy+yz+zx) +9$ which is true because $(x+y+z)^2\geq 9$ and $(x+y+z)^2\geq 3(xy+yz+zx)$

From Cauch -Schwarz inequality we have; $((xy+1)+(yz+1)+(xz+1))(\frac{1}{2}+ \frac{1}{2}+ \frac{1}{2})\geq (RHS)^2$ If we prove $(x+y+z)^2\geq \frac{3}{2}(xy+yz+zx+3)$ we’ll be done Let $x+y+z=u$ and $xy+yz+zx=v$ if $u^2\geq \frac{3}{2}(v+3)$ then $(1)u^2\geq 3v$ $(2) u^2\geq 9$ adding these inequalities gives the desired result Who is the author btw?