Prove that $p(x^q-1) + q(x^{-p}-1) \geq 0$ under certain conditions

Question

Question: Prove that $p(x^q-1) + q(x^{-p}-1) \geq 0$ for $x > 0$ and $p,q \in \mathbb{Z}^+$.

My attempts: I initially thought to make the expression $px^q + qx^{-p} \geq p+q$ and see if that helped, although I'm not sure where to go from here. Another idea was to multiply the LHS by $x^p$ to give $px^{p+q} + q + (p-q)x^p$. All I would need to do here is prove that this is greater than $0$ but I'm not entirely sure if it's reasonable to. Perhaps I'm missing an inequality identity that I could use, however for the meantime I'm stuck. Any help or guidance would be greatly appreciated!

Answer 1

Use AM-GM inequality$$px^q+\frac{q}{x^p}=\underbrace{x^q+x^q+\cdots+x^q}_{p\text{ times}}+\underbrace{\frac1{x^p}+\frac1{x^p}+\cdots+\frac1{x^p}}_{q\text{ times}} \ge p+q\\$$ $$\implies px^q-p =qx^{-p}-q$$

EDIT
This can be extended to the case where $p,q\in \mathbb{R^+}$. When $p,q\in \mathbb{R^+}$ we can use weighted AM-GM inequality

Answer 2

Let $$f(x)=\frac{1}{q}x^q+\frac{1}{p}x^{-p}-\frac1p-\frac1q$$

Then $$f’(x)=x^{q-1}-x^{-p-1}=x^{-p-1}(x^{p+q}-1)$$ and $f’(x)=0$ at $x=1$ and $f’(x)<0$ when $x<1$ and $f’(x)>0$ when $x>1?$ So the minimum of $f$ is $f(1)=0$

Therefore:

$$\frac{1}{q}x^q+\frac{1}{p}x^{-p}\geq\frac1p+\frac1q$$

Multiply both sides by $pq.$