# If $xy+xz+yz=1+2xyz$ then $\sqrt{x}+\sqrt{y}+\sqrt{z}\geq2$.

Let $x$, $y$ and $z$ be non-negative numbers such that $xy+xz+yz=1+2xyz$. Prove that: $$\sqrt{x}+\sqrt{y}+\sqrt{z}\geq2$$

The equality occurs for $x=y=1$ and $z=0$.

I tried Lagrange Multipliers and more, but I don't see a proof.

• If $x,y,z>0$ than $$(1+2xyz)^2=(xy+yz+zx)^2 \ge 3xyz(x+y+z)$$ So $$x+y+z \le \frac{(1+2xyz)^2}{3xyz}$$ Also $$(x+y+z)^2 \ge 3xy+yz+zx=3+6xyz$$ Feb 3, 2017 at 15:35
• @S.C.B., did you mean $3(xy+yz+zx)$ in the last line of your comment? Feb 3, 2017 at 15:43
• @BarryCipra Yep, typo. Feb 3, 2017 at 15:43
• @MichaelRozenberg Are you typing an answer? What is the basic idea? Feb 3, 2017 at 15:50
• @MichaelRozenberg So you're not typing an answer? Feb 3, 2017 at 15:52

Let: $x=a^2 , y=b^2 , z=c^2$

So we must prove : $a+b+c \ge 2$

for nonnegative $a, b, c : \ a^2b^2+b^2c^2+c^2a^2=1+2a^2b^2c^2$

$$p=a+b+c\ , \ q=ab+bc+ca \ , \ r=abc$$

$$q^2=1+2pr+2r^2 \Rightarrow q \ge 1$$

1. $q \ge \dfrac{4}{3} \Rightarrow p^2 \ge 3q \ge 4$

2. $1 \le q \le \dfrac{4}{3}\ , \ 2pr = q^2-1-2r^2 \le q^2-1$

By Shur we have : $p^3+9r \ge 4pq \Rightarrow p^4-4qp^2+\dfrac{9}{2}(q^2-1)\ge 0\Rightarrow$

$$p^2\ge 2q+\sqrt{\dfrac{9-q^2}{2}} \ge 4 \Leftrightarrow (q-1)(23-9q)\ge 0$$

Equality holdes for : $a=b=1 \ , \ c=0$

• Nice! Also there is a nice solution because we can assume that alls $xy\leq1$ and from here $\sum\limits_{cyc}\sqrt{xy}\geq\sum\limits_{cyc}xy$. Mar 14, 2017 at 21:38

Short proof.

Clearly $$xy+yz+zx \ge 1$$

We have by AM-GM

$$(\sqrt{x}+\sqrt{y}+\sqrt{z})^2=x+y+z+2(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})$$

$$\ge x+y+z+ \frac{4xy}{x+y}+\frac{4yz}{y+z}+\frac{4zx}{z+x}$$

$$\ge x+y+z + \frac{4(xy+yz+zx)}{x+y+z} \ge x+y+z+\frac{4}{x+y+z} \ge 4$$

The proof is complete.