By Kiran Kedlaya
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For a simple example, let f (x, y, z) = ax + by + cz with a, b, c constants, not all zero, and consider the constraint g(x, y, z) = 1, where g(x, y, z) = x2 + y 2 + z 2 . Then the Lagrange multiplier condition is that a = 2λx, b = 2λy, c = 2λz. The only points satisfying this condition plus the original constraint are ±√ a2 1 (a, b, c), + b2 + c 2 and these are indeed the minimum and maximum for f under the constraint, as you may verify geometrically. 1 Quick reference Here’s a handy reference guide to the techniques we’ve introduced.
5 1. (IMO 1968/2) Prove that for all real numbers x1 , x2 , y1 , y2 , z1 , z2 with x1 , x2 > 0 and x1 y1 > z12 , x2 y2 > z2 , the inequality 8 1 1 ≤ + 2 2 (x1 + x2 )(y1 + y2 ) − (z1 + z2 ) x1 y1 − z1 x2 y2 − z22 is satisfied, and determine when equality holds. (Yes, you really can apply the material of this section to the IMO! 6 Constrained extrema and Lagrange multipliers In the multivariable realm, a new phenomenon emerges that we did not have to consider in the one-dimensional case: sometimes we are asked to prove an inequality in the case where the variables satisfy some constraint.
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.
A < B by Kiran Kedlaya