let f(x) = 4x^4 - 2x + 1. f'(x) = 16x^3 - 2 = 2(2x - 1)(4x^2 + 2x + 1). f(x) decreases in x < 1/2, and increases in x > 1/2. For any real number x, f(x) ≥ f(1/2) = 1/4.
(a + b + c) + (1/a + 1/b + 1/c) = (a + 1/a) + (b + 1/b) + (c + 1/c) ≥ 2√(a*(1/a)) + 2√(b*(1/b)) + 2√(c*(1/c)) = 6. This result tells that either (a + b + c) or (1/a + 1/b + 1/c) is greater than or equal to 3. If both of them were smaller than 3, the sum of them would be less than 6. Note that your proposition is FALSE. When a = b = c = 1, neither (a + b + c) or (1/a + 1/b + 1/c) is equal to 3, and not greater than 3.
>>39 Can you translate that? I guess your answer is right,and the Japanese answer that is witten first is wrong.
>>40 Please tell me how you got to speak English. Are you Japanese ?
He can read the problem written in Japanese, which means he understands Japanese, you dork!
>>53 You're not a Japanese,are you?If so,let me know why you could understand the problem written in Japanese. I was born in America and had lived there for 4 years.(Now I'm in Japan.)That's why I can speak English,but I think most junior high school students in Japan can speak English as well as me.