Question 109
Let $x, y, z$ be nonnegative real numbers with $x y + y z + z x = 1$. Show that
$$
\frac{4}{x+y+z} \leq (x+y)(\sqrt{3} z + 1)
$$
$$
\frac{4}{x+y+z} \leq (x+y)(\sqrt{3} z + 1)
$$
Math Olympiad Vault · MathNet 2017-part-06 — 176 problems (EN–VI)