PROVE THATZ 1Z 1DX DYX −1 + | LN Y| − 1 ≤ 1. [20 POINTS]00SOLUTION...

5. Prove that

Z 1

dx dy

x ⁻¹ + | ln y| − 1 ≤ 1. [20 points]

0

Solution 1. First we use the inequality

x ⁻¹ − 1 ≥ | ln x|, x ∈ (0, 1],

which follows from

(x ⁻¹ − 1)

x=1 = | ln x|| _x=1 = 0,

(x ⁻¹ − 1) ⁰ = − 1

x ² ≤ − 1

x = | ln x| ⁰ , x ∈ (0, 1].

Therefore

| ln x| + | ln y| =

| ln(x · y)| .

x ⁻¹ + | ln y| − 1 ≤

Substituting y = u/x, we obtain

du

dx

| ln u| · du

| ln(x · y)| =

| ln u| = 1.

| ln u| =

x

u

Solution 2. Substituting s = x ⁻¹ − 1 and u = s − ln y,

e ^−u

Z u

Z ∞

e ^s

e ^s−u

u dsdu.

(s + 1) ² ds

(s + 1) ² u duds =

x ⁻¹ + | ln y| − 1 =

s

Since the function _(s+1) ^e

is convex,

e ^u

(u + 1) ² + 1

2

(s + 1) ² ds ≤ u

so

u

du

e ^−u du

= 1.

u du = 1

(u + 1) ² +