How do I show $x^x > k \cdot a^x $?

Question

If I know that $x \geq ka$, how do I get to $x^x > k \cdot a^x $ and obtain a strict inequality? Here $ a,x \in \mathbb{N}$ and $k \in \mathbb{R^+}$.

Answer 1

We have that

• for $k\ge 1$ as $x>ka>1$

$$x^x >(ka)^{ka}=k^{ka}a^{ka}>ka^x $$

• for $0<k<1$ as $x>a$

$$x^x >a^a>ka^x $$

Answer 2

Ask yourself what happens if $x=1,k=1,a=1$?

Your claim becomes wrong. So just assume $x > ka$.

What happens if now $x=2, a=4, k=0.49$?

It's a slightly bit more complicated.

If $a=0$, there is nothing left to prove since $x > 0$.

The crucial inequality is

$$ x^x > k a^x$$

If we rewrite that as the equivalent

$$ \left(\frac{x}a\right)^x > k$$

It should be easy to see that we can choose $x=\max(2ka,a)$. Since $\frac{x}a \ge 1$ and $x \ge a \ge 1$, we have $ \left(\frac{x}a\right)^x \ge \left(\frac{x}a\right)^1=\frac{x}a \ge 2k > k.$

Of course, you can replace $2ka$ with any $(1+\epsilon)ka$.