2
$\begingroup$

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....

$\endgroup$
  • 4
    $\begingroup$ Can you please include parentheses in the appropriate places? It's a bit difficult to understand what goes where. $\endgroup$ – Cameron Williams Feb 23 '14 at 18:05
  • $\begingroup$ Are the 1993's part of the argument to $\log$ or just a constant multiplying everything? $\endgroup$ – copper.hat Feb 23 '14 at 18:11
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.