Final answer:
To prove (x²+y²)¹/² ≥ (x³+y³)¹/³ for any x>0 and y>0, introduce z=(xy)² and apply the binomial theorem to simplify the inequality.
Step-by-step explanation:
To prove that (x²+y²)¹/² ≥ (x³+y³)¹/³ holds for any x>0 and y>0, we can introduce a new variable z=(xy)². By expanding (x²+y²)¹/² and (x³+y³)¹/³ using the binomial theorem, we can show that (x²+y²)¹/² ≥ (x³+y³)¹/³ simplifies to z³ ≥ 2xy². We can further simplify this inequality by substituting z=(xy)², which gives us (xy)⁶ ≥ 2(xy)². Cancelling out xy from both sides, we get (xy)⁴ ≥ 2. Since xy>0, it follows that xy ≥ √2. Therefore, (x²+y²)¹/² ≥ (x³+y³)¹/³ is true for any x>0 and y>0.