Vrftjkry

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.

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?