Showing that a function is increasing and then proving an inequality

Question

Let $f:[0, \infty) \to \mathbb{R} $, $f(x) =3^x+10^x+21^x-6^x-7^x-15^x$. Prove that :
a) $f$ is an increasing function.
b) $2\cdot 3^x+9\cdot 10^x+20\cdot 21^x\ge 5\cdot 6^x+6\cdot 7^x+14\cdot 15^x$,$\forall x\ge 0$
I tried differentiating the function, but it didn't help. Then I thought about factoring the function, but this didn't work either.

Answer 1

We can write $$ f(x) = (3^x-2^x)(7^x-5^x)+(2^x-1)(7^x-3^x). $$ It can be easily seen that each factor is a non-negative, increasing function, for example, $$ 3^x-2^x = 2^x\left( \left (\frac{3}{2}\right)^x-1\right) $$ is a product of two non-negative, increasing functions. Hence $f$ is also an incresing function.

The second assertion is equivalent to $f(x+1)-f(x)\ge 0$. This is true since $f$ is increasing.