Prove that $\log_8(9)+\log_9(10)+\log(11)<2\log_2(3)$ Original title had a typo, third term of the LHS is $\log(11)$.
Prove that $\log_8(9)+\log_9(10)+\log(11)<2\log_2(3)$
I am kind of frustrated with this simple problem. How do you prove this without using a calculator. I know that both LHS and RHS are greater than 3. On the left hand side, $\log_2(9)>\log_2(8)=3$. On the RHS, each term should be slighly greater than 1. How should I start?

Lemma: For any natural number $n\ge2$: 
$$\log_n(n+1) \gt \log_{n+1}(n+2)\qquad(1)$$
Proof: Check the last inequality proved on the following page: Logarithmic ineqaulities
Now transform the inequality in the following way:
$$\log_89+\log_910 +\log_{10}11<2\log_23$$
$$\log_89+\log_910 +\log_{10}11<\log_23^2$$
$$\log_89+\log_910 +\log_{10}11<\log_{2^3}3^6$$
$$\log_89+\log_910 +\log_{10}11<\log_{8}(9\cdot9^2)$$
$$\log_89+\log_910 +\log_{10}11<\log_{8}9 + \log_89^2$$
$$\log_910 +\log_{10}11<2\log_89$$
which is obviously true, because lemma (1) guarantees that:
$$\log_89 \gt \log_910 \gt \log_{10}11$$ 
