Pinko wrote: $x$, $y$, and $z$ are positive real numbers satisfying the equation $x+y+z=\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$. Prove the following inequality: $xy + yz + zx \geq 3$. Put $\sigma_1=x+y+z$, $\sigma_2=xy+yz+xz$ and $\sigma_3=xyz$. Then $\sigma_2=\sigma-1\sigma_3$. On the other hand, $$\sigma_2^2\geq 3\sigma_1\sigma_3=3\sigma_2,$$so $\sigma_2\geq 3$.

If $x$, $y$, and $z$ are positive real numbers satisfying the equation $x+y+z=\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$. Prove the following inequality: $x+y+z+\frac{3}{x+y+z} \ge 4xyz \ \ ; $. Greetings!

Pinko wrote: $x$, $y$, and $z$ are positive real numbers satisfying the equation $x+y+z=\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$. Prove the following inequality: $xy + yz + zx \geq 3$. Proof of Zhangyunhua:

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Lonesan wrote: If $x$, $y$, and $z$ are positive real numbers satisfying the equation $x+y+z=\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$. Prove the following inequality: $x+y+z+\frac{3}{x+y+z} \ge 4xyz \ \ ; $. Greetings! Equivalent with the following one: $ 2(xy+yz+zx)-(x^2 +y^2 +z^2) \le 3 \ \ ; $ Greetings from Lorian Saceanu

If $x$, $y$, and $z$ are positive real numbers satisfying the equation $x+y+z=\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$. $ \frac{x}{y} + \frac{y}{z} + \frac{z}{x} \ge x+y+z \ \ ; $ Greetings!