# Let $x_0 > x_1 > x_2>x_3$ be any positive real numbers . What is the largest value of the real number k such that..

Question :

Let $x_0 > x_1 > x_2>x_3$ be any positive real numbers . What is the largest value of the real number k such that $$\log \frac{x_0}{x_1}1993 + \log \frac{x_1}{x_2}1993 +\log \frac{x_2}{x_3} 1993 \geq k \log \frac{\log x_0 }{x_3}1993$$

How to solve such problems, please suggest thanks....

• Can you please include parentheses in the appropriate places? It's a bit difficult to understand what goes where. – Cameron Williams Feb 23 '14 at 18:05
• Are the 1993's part of the argument to $\log$ or just a constant multiplying everything? – copper.hat Feb 23 '14 at 18:11

Let $y_i =log_{1993} \frac{x_i-1}{x_i}$

for $i=1,2,3.$ Then the given inequality may be written as

$$\frac{1}{y_1} +\frac{1}{y_2} +\frac{1}{y_3} \geq \frac{k}{y_1 + y_2 +y_3}$$

By using arihtmetic - geometric mean inequality :

$(\frac{1}{y_1} +\frac{1}{y_2} +\frac{1}{y_3} ) (y_1+y_2+y_3) \geq 3(\sqrt[3]{\frac{1}{y_1 y_2 y_3}})3(\sqrt[3]{y_1 y_2 y_3}) = 9$

Equality holds if and only if $y_1 =y_2 =y_3$ or $x_0 ,x_1,x_2,x_3$ forms a geometric progression .

Hence maximum value of k is 9.