a, b, c are three positive numbers, no two of which are equal. Prove that a^2 - ab + b^2 > ab a^3 + b^3 > ab (a + b) 2(a^3 + b^3 + c^3) > ab (a + b) + bc (b + c) + ca (c + a)
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Originally Posted by Detanon a, b, c are three positive numbers, no two of which are equal. Prove that a^2 - ab + b^2 > ab This is equivalent to a^3 + b^3 > ab (a + b) Multiply the preceding inequality by 2(a^3 + b^3 + c^3) > ab (a + b) + bc (b + c) + ca (c + a) From the preceding, (Just substituting different variables) Now add.
Why is ?? Is it because its a perfect square or something so its greater than zero? I dont really get the second steps either.. what did you do with a^3 + b^3 > ab (a + b)
Originally Posted by Detanon Why is ?? Is it because its a perfect square or something so its greater than zero? The square of any non-zero number is positive. We know because we were told . I dont really get the second steps either.. what did you do with a^3 + b^3 > ab (a + b) Multiply by . What do you get?
a^3 + b^3? But why do you multiply a^2 -ab +b^2 by a+b. Can you show all the steps please.
Originally Posted by Detanon a^3 + b^3? But why do you multiply a^2 -ab +b^2 by a+b. Can you show all the steps please. I have listed all the steps. I suggest you go back and re-read my first post.
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