404
CHAPTER 43
Fig. 43-6
43.68 Prove Cauchy's inequality:
Let
and
Let x, = aJA and y^bJB. (If A=0 or B = 0,
the desired result is obvious.) Then
and
By Problem 43.66,
Hence,
substitution in the equation for 5.)
Then sina=V5/3, csca=3/V5, cota=2/V5, cot a-esc a =-1/V5. Therefore, -r/V5 = r-A.
(The value of r, and therefore of h, can be found in terms of S by
Then
Hence, by (2), r = 2A. In addition, 2rh + r
2 cot a = 2A(r + h + r esc a) follows from (1) and yields, with
r = 2A, 2rh + r
2 cot a = r(r + h + rcsca), and, therefore, rVcot a - esc a) = iir - h). Hence, r(cot acsca) = r-h. From (3) and r = 2A, we get
esc a cot a, from which follows cos
esc
CHAPTER 43
Fig. 43-6
43.68 Prove Cauchy's inequality:
Let
and
Let x, = aJA and y^bJB. (If A=0 or B = 0,
the desired result is obvious.) Then
and
By Problem 43.66,
Hence,
substitution in the equation for 5.)
Then sina=V5/3, csca=3/V5, cota=2/V5, cot a-esc a =-1/V5. Therefore, -r/V5 = r-A.
(The value of r, and therefore of h, can be found in terms of S by
Then
Hence, by (2), r = 2A. In addition, 2rh + r
2 cot a = 2A(r + h + r esc a) follows from (1) and yields, with
r = 2A, 2rh + r
2 cot a = r(r + h + rcsca), and, therefore, rVcot a - esc a) = iir - h). Hence, r(cot acsca) = r-h. From (3) and r = 2A, we get
esc a cot a, from which follows cos
esc