How do we get to the inequality?

[evinda]

Hello! (Wave)Given that $n \geq 15$, how can we conclude the following? (Thinking)

$$cn \lg n- cn \lg \left ( \frac{3}{2}\right )+\frac{n}{2}+15 c \lg n-15 c \lg \left ( \frac{3}{2}\right) \leq cn \lg n, \text{ for } c>1 \text{ and } n>15$$

 

[Ackbach]

I take it $\text{lg}\equiv \log_{10}$, so that you are desiring to prove

$$cn \log_{10}(n)- cn \log_{10} \left ( \frac{3}{2}\right )+\frac{n}{2}+15 c \log_{10}(n)-15 c \log_{10} \left ( \frac{3}{2}\right) \leq cn \log_{10}(n), \text{ for } c>1 \text{ and } n>15.$$

We may subtract $c n \log_{10}(n)$ from both sides to obtain the equivalent inequality

$$- cn \log_{10} \left ( \frac{3}{2}\right )+\frac{n}{2}+15 c \log_{10}(n)-15 c \log_{10} \left ( \frac{3}{2}\right) \leq 0, \text{ for } c>1 \text{ and } n>15.$$

Choosing $c=2$ and $n=16$, we find that the LHS of this inequality is approximately $16.6 \not\le 0$. Therefore, the inequality is false.