Problem - 3223
Let $a, b, c$ be positive real numbers. Show that $$6a+4b+5c\ge 5\sqrt{ab} + 3\sqrt{bc} + 7\sqrt{ca}$$
By AM-GM, we have
$$5\sqrt{ab} + 3\sqrt{bc} + 7\sqrt{ca}\le 5\Big(\frac{a+b}{2}\Big)+3\Big(\frac{b+c}{2}\Big)+7\Big(\frac{c+a}{2}\Big)=6a+4b+5c$$