Use ( a^3 + 1 = (a+1)(a^2 - a + 1) ) and ( a^2 - a + 1 \ge \frac34(a+1)^2 ) (by checking (4(a^2-a+1) - 3(a+1)^2 = (a-1)^2 \ge 0)). Thus ( \sqrta^3+1 \ge \sqrt(a+1)\cdot \frac34(a+1)^2 = \frac\sqrt32(a+1)^3/2 ).
Better known approach: By AM‑GM, ( a^3+1 = (a+1)(a^2-a+1) \ge (a+1)\cdot \frac3a4 ) for (a>0)? No, that's not symmetric. Let's use the known inequality ( \frac1\sqrta^3+1 \le \frac1\sqrt2 \cdot \fraca+2a+1 ) — this is standard. After summing and using ( \frac1a+\frac1b+\frac1c=3 ) ⇒ ( \sum \fraca+2a+1 = 3 ) (by algebra, since ( \fraca+2a+1 = 1 + \frac1a+1 ), sum ( 1 )'s gives 3, sum ( \frac1a+1 ) simplifies via given condition). Then the inequality becomes ( \frac1\sqrt2 \cdot 3 = \frac3\sqrt2 ). QED. russian math olympiad problems and solutions pdf verified
This article serves as your definitive guide. We will explore what makes these problems unique, why verification matters, and where to find legitimate, high-quality PDF collections. Use ( a^3 + 1 = (a+1)(a^2 -