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The question requires us to prove 2x³+ 1/27≥x², but in reality, we only need to demonstrate that 2x³-x²+1/27≥0, which holds true after rearranging terms. So we first assume f(x)=2x³-x²+1/27. Next, we aim to prove that when x≥0, the minimum value of the function is ≥0, indicating that f(x)=2x³-x²+1/27 is ≥0 for x≥0. After rearranging, 2x³+1/27 is then ≥x²
I see! ! I understand! ! Thank you for your detailed explanations each time. It really helps me every time😊
Sorry for all the questions😭
I understand that the minimum value is 0 when x = 1/3, but I don't understand why 2 x^3 + 1/27≧ x^2 holds from there.