4.2 Cauchy–Schwarz Revisited
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Problem 1.25 Let x, y, z > 0. Prove that y z 1 x + + ≥ . x + 2y + 3z y + 2z + 3x z + 2x + 3y 2 Solution We write the left-hand side as y2 z2 x2 + + x 2 + 2xy + 3xz y 2 + 2yz + 3xy z2 + 2xz + 3yz and apply the lemma. This yields y z (x + y + z)2 x + + ≥ 2 . x + 2y + 3z y + 2z + 3x z + 2x + 3y x + y 2 + z2 + 5(xy + xz + yz) Now it suffices to prove that
x2
+ y2
1 (x + y + z)2 ≥ , 2 + z + 5(xy + xz + yz) 2
but this is equivalent to x 2 + y 2 + z2 ≥ xy + xz + yz, proved in Problem 1.19. Problem 1.26 Let x, y, z > 0. Prove that x2 y2 z2 3 + + ≥ . (x + y)(x + z) (y + z)(y + x) (z + x)(z + y) 4 Solution We apply the lemma again to obtain y2 z2 x2 + + (x + y)(x + z) (y + z)(y + x) (z + x)(z + y) ≥
(x + y + z)2 . x 2 + y 2 + z2 + 3(xy + xz + yz)
The inequality x2
+ y2
(x + y + z)2 3 ≥ 2 + z + 3(xy + xz + yz) 4
is equivalent to x 2 + y 2 + z2 ≥ xy + xz + yz.