Given $n_1 \ge d_1 \ge 0$, $n_2 \ge d_2 \ge 0$ and $d_1 + d_2 \gt 0$, how do I show that...

Question

Given $n_1 \ge d_1 \ge 0$, $n_2 \ge d_2 \ge 0$ and $d_1 + d_2 \gt 0$, how do I show that $$\Big( \frac{n_1}{d_1} \Big)^{d_1} \Big( \frac{n_2 }{d_2} \Big)^{d_2} \le \Big(\frac{(n_1 + n_2)}{d_1 + d_2}\Big)^{d_1}\Big(\frac{(n_1 + n_2)}{d_1 + d_2}\Big)^{d_2}$$

I have managed to split this into 3 cases:

Case 1: $n_1d_2 = d_1n_2$

This leads to $\Big( \frac{n_1}{d_1} \Big)^{d_1} = \Big(\frac{(n_1 + n_2)}{d_1 + d_2}\Big)^{d_1}$, and also leads to $\Big( \frac{n_2}{d_2} \Big)^{d_2} = \Big(\frac{(n_1 + n_2)}{d_1 + d_2}\Big)^{d_2}$, which satisfies the expression above.

My problem here is with cases 2 and 3.

For example, Case 2: $n_1d_2 \le n_2d_1$

This leads to $\Big( \frac{n_1}{d_1} \Big)^{d_1} \le \Big(\frac{(n_1 + n_2)}{d_1 + d_2}\Big)^{d_1}$, and also leads to $\Big( \frac{n_2}{d_2} \Big)^{d_2} \ge \Big(\frac{(n_1 + n_2)}{d_1 + d_2}\Big)^{d_2}$, which's where I am conflicted.

How can I show that this conflict works itself out to satisfy th main expression?

Thank you.

Answer 1

You can show your inequality by taking the logarithm and using the concavity of $\log$:

• $\log(\lambda x + (1-\lambda)y) \geq \lambda \log x + (1-\lambda) \log y$ for $\lambda \in [0,1]$

Set

• $\lambda = \frac{d_1}{d_1+d_2} \Rightarrow 1-\lambda = \frac{d_2}{d_1+d_2}$
• $x= \frac{n_1}{d_1}$
• $y= \frac{n_2}{d_2}$

Taking the logarithm on the right side of your inequality gives

$$d_1 \log \frac{n_1}{d_1} + d_2 \log \frac{n_2}{d_2} \le (d_1+d_2) \log \left( \frac{n_1}{d_1+d_2} + \frac{n_2}{d_1+d_2}\right)$$

Can you take it from here?