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By Kiran Kedlaya

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Mini Habits: Smaller Habits, Bigger Results

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Like most folk who attempt to swap and fail, i thought that i used to be the problem.

Then one afternoon–after one other failed try to get prompted to exercise–I (accidentally) began my first mini behavior. I firstly devoted to do one push-up, and it became a whole exercise routine. i used to be stunned. This "stupid idea" wasn't purported to paintings. i used to be stunned back while my good fortune with this method persisted for months (and to this day). I needed to think of that perhaps I wasn't the matter in these 10 years of mediocre effects. perhaps it used to be my previous innovations that have been useless, regardless of being oft-repeated as "the method to change" in numerous books and blogs.

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Prove that 1 x3 + y 3 + z 3 + 6xyz ≥ . 4 34 5. (Taiwan, 1995) Let P (x) = 1 + a1 x + · · · + an−1 xn−1 + xn be a polynomial with complex coefficients. Suppose the roots of P (x) are α1 , α2 , . . , αn with |α1 | > 1, |α2 | > 1, . . , |αj | > 1 and |αj+1 | ≤ 1, |αj+2 | ≤ 1, . . , |αn | ≤ 1. Prove that j |αi | ≤ |a0 |2 + |a1 |2 + · · · + |an |2 . ) 6. Prove that, for any real numbers x, y, z, 3(x2 − x + 1)(y 2 − y + 1)(z 2 − z + 1) ≥ (xyz)2 + xyz + 1. 7. (a) Prove that any polynomial P (x) such that P (x) ≥ 0 for all real x can be written as the sum of the squares of two polynomials.

Bucharest 40 (1934), 155-160. [4] S. Rabinowitz, Index to Mathematical Problems 1980-1984, Mathpro Press, Westford (Massachusetts), 1992.

6. Prove that, for any real numbers x, y, z, 3(x2 − x + 1)(y 2 − y + 1)(z 2 − z + 1) ≥ (xyz)2 + xyz + 1. 7. (a) Prove that any polynomial P (x) such that P (x) ≥ 0 for all real x can be written as the sum of the squares of two polynomials. (b) Prove that the polynomial x2 (x2 − y 2 )(x2 − 1) + y 2 (y 2 − 1)(y 2 − x2 ) + (1 − x2 )(1 − y 2 ) is everywhere nonnegative, but cannot be written as the sum of squares of any number of polynomials. S. Bullen, D. Vasi´c, Means and their Inequalities, Reidel, Dordrecht, 1988.

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