1. ## logarithms

Given two real numbers a and b where a>1 and b>1, prove that
$\displaystyle \frac{1}{\log_b a}+\frac{1}{\log_a b}\geq 2$

Attempt:

$\displaystyle (\log_a b)(\log_b a)=1$

$\displaystyle \frac{1}{\log_b a}+\frac{1}{\log_a b}$

$\displaystyle =\frac{\log_a b+\log_b a}{(\log_a b)(\log_b a)}$

I have to prove that the numerator is >=2 but i just don see a way..

2. Originally Posted by hooke
Given two real numbers a and b where a>1 and b>1, prove that
$\displaystyle \frac{1}{\log_b a}+\frac{1}{\log_a b}\geq 2$

Attempt:

$\displaystyle (\log_a b)(\log_b a)=1$

...
1. Re-write the given inequality:

$\displaystyle \frac{1}{\log_b( a)}+\log_b(a)\geq 2$

2. Since a > 1 and b > 1 the logarithms are greater than zero, that means they are positive. Multiply the inequality by $\displaystyle \log_b(a)$:

$\displaystyle 1+\left(\log_b(a)\right)^2\geq 2 \cdot \log_b(a)$

3. You'll get:

$\displaystyle \left(\log_b(a)\right)^2- 2 \cdot \log_b(a) +1 \geq 0$

$\displaystyle \left(\log_b(a)-1\right)^2\geq 0$

4. A square is allways greater or equal zero.

3. Originally Posted by earboth
1. Re-write the given inequality:

$\displaystyle \frac{1}{\log_b( a)}+\log_b(a)\geq 2$

2. Since a > 1 and b > 1 the logarithms are greater than zero, that means they are positive. Multiply the inequality by $\displaystyle \log_b(a)$:

$\displaystyle 1+\left(\log_b(a)\right)^2\geq 2 \cdot \log_b(a)$

3. You'll get:

$\displaystyle \left(\log_b(a)\right)^2- 2 \cdot \log_b(a) +1 \geq 0$

$\displaystyle \left(\log_b(a)-1\right)^2\geq 0$

4. A square is allways greater or equal zero.
Thanks earboth, i thought of this too but after pondering a while, shouldn't we prove from any side until we get the result instead of proving that the result is true.

Take trigonometric identities for example,

I am supposed to prove from either side so that it equals the other side.