HKHLE 1991 Math Question

a) Differentiating the given eqn wrt t:

rur-1u' = ryr-1y' + rzr-1z'

ur-1u' = yr-1y' + zr-1z' ... (*)

u' = (y/u)r-1y' + (z/u)r-1z'

Diff (*) wrt t:

ur-1u" + (r - 1)ur-2(u')2 = yr-1y" + (r - 1)yr-2(y')2 + zr-1z" + (r - 1)zr-2(z')2

ur-1u" + (r - 1)ur-2[(y/u)r-1y' + (z/u)r-1z']2 = (r - 1)yr-2(y')2 + (r - 1)zr-2(z')2

ur-1u" + (r - 1)(1/u)r(yr-1y' + yr-1z')2 = (r - 1)yr-2(y')2 + (r - 1)zr-2(z')2

Multiplying ur to both sides:

u2r-1u" + (r - 1)(yr-1y' + yr-1z')2 = (r - 1)ur[yr-2(y')2 + zr-2(z')2]

u2r-1u" = (r - 1)(yr + zr)[yr-2(y')2 + zr-2(z')2] - (r - 1)(yr-1y' + yr-1z')2

= (r - 1){(yr + zr)[yr-2(y')2 + zr-2(z')2] - (yr-1y' + yr-1z')2}

b) Obviously, equality holds when x = x0

Consider the expression:

[f(x) - f(x0)]/(x - x0)

When x < x0, by MVT, there exists x < a < x0 s.t.

f'(a) = [f(x) - f(x0)]/(x - x0)

So [f(x) - f(x0)]/(x - x0) < f'(x0) since f'(x) in increasing

f(x) - f(x0) > f'(x0) since x - x0 < 0

When x > x0, by MVT, there exists x0 < b < x s.t.

f'(b) = [f(x) - f(x0)]/(x - x0)

So [f(x) - f(x0)]/(x - x0) > f'(x0) since f'(x) in increasing

f(x) - f(x0) > f'(x0) since x - x0 > 0

So conclusively we have f(x) > f(x0) + f'(x0)(x - x0)

c1) Note that y" = z" = 0 for all 0 <= t <= 1. So by (a):

u2r-1u" = (r - 1){(yr + zr)[yr-2(y')2 + zr-2(z')2] - (yr-1y' + yr-1z')2}

Then by the CS ineq:

(c12 + c22)(d12 + d22) >= (c1d1 + c2d2)2

Sub c1 = yr/2, c2 = zr/2, d1 = y(r-2)/2y' and d2 = z(r-2)/2z':

(yr + zr)[yr-2(y')2 + zr-2(z')2] >= (yr-1y' + zr-1z')2

(yr + zr)[yr-2(y')2 + zr-2(z')2] - (yr-1y' + yr-1z')2 >= 0

u" >= 0 since r > 1 and u > 0

c2) When t = 0, u = (b1r + b2r)1/r and when t = 1, u = (a1r + a2r)1/r.

When t = 1/2, u = [(a1 + b1)r/2r + (a2 + b2)r/2r]1/r = (1/2)[(a1 + b1)r + (a2 + b2)r]1/r

So by (c1) and (b):

(1/2)[(a1 + b1)r + (a2 + b2)r]1/r <= (1/2)[(a1r + a2r)1/r + (b1r + b2r)1/r]

[(a1 + b1)r + (a2 + b2)r]1/r <= [(a1r + a2r)1/r + (b1r + b2r)1/r]