How to prove that: $\log_{{1\over 2}}(3) + \log_3\left({1 \over 2}\right) < -2$

Question

Prove that:

$$\log_{{1\over 2}}(3) + \log_3\left({1 \over 2}\right) < -2$$

Please help me solve it.

Answer 1

If you know the AM-GM inequality, this is easy.

$a=\log_{\frac{1}{2}}(3)=-\log_23$ and $b=\log_3(\dfrac{1}{2})=-\log_32$.

Now $-a>0,-b>0$ so $-a-b>2\sqrt{|ab|}=2$ implying $a+b<-2$.

Answer 2

$$\log_{3}(1/2)+\log_{1/2}(3)=\log_3(1/2)+\frac{\log_3(3)}{\log_3(1/2)}=$$

$$=\log_{3}(1/2)+\frac{1}{\log_3(1/2)}$$

$$=-\left(\left(\sqrt{-\log_{3}(1/2)}-\sqrt{-\frac{1}{\log_3(1/2)}}\right)^2+2\right)\le -2$$

Equality can't occur, because $\sqrt{-\log_{3}(1/2)}\neq \sqrt{-\frac{1}{\log_3(1/2)}}$.

Expressed differently:

$$\left(-\log_{3}(1/2)\right)+\left(-\log_{1/2}(3)\right)=\left(-\log_{3}(1/2)\right)+\left(-\frac{1}{\log_3(1/2)}\right)$$

$$\ge 2\sqrt{\left(-\log_3(1/2)\right)\left(-\frac{1}{\log_3(1/2)}\right)}=2$$

You may call it AM-GM, but it's the trivial inequality $a+b\ge 2\sqrt{ab}$, i.e. $\left(\sqrt{a}-\sqrt{b}\right)^2\ge 0$, for $a,b\ge 0$. Equality won't occur, since $\left(-\log_3(1/2)\right)\neq -\frac{1}{\log_3(1/2)}$.

Answer 3

Let $ a=\log_{3}\frac{1}{2}$, so $\;a<0$ and $\;\displaystyle\log_{\frac{1}{2}}3=\frac{\log_{3}3}{\log_3\frac{1}{2}}=\frac{1}{a}$.

Then $(a+1)^2>0\implies a^2+1>-2a\implies a+\frac{1}{a}<-2\;\;$ (since $a<0$)