minimum value of $ f(x) $ without the use of derivative

Question

Minimum value of $\displaystyle f(x) = \frac{x^p}{p}+\frac{x^{-q}}{q}$ subjected to $\displaystyle \frac{1}{p}+\frac{1}{q} = 1$ and $p>1$

Although I have solved it using derivative. But did not understand how can i solve

without derivative, Help Required, Thanks

Answer 1

Use Young's inequality. $$ f(x)=\frac{x^p}{p}+\frac{x^{-q}}{q}\ge x\cdot\frac{1}{x}=1. $$ Since $f(1)=1$, the minimum is $1$.

Answer 2

By AM-GM: $$\frac{x^p}{p}+\frac{x^{-q}}{q}\geq\frac{1}{p}+\frac{1}{q}+x+\frac{1}{x}-2\geq\frac{1}{p}+\frac{1}{q}=1$$

Since $x^p-px+p-1\geq0$ and $x^{-q}-\frac{q}{x}+q-1\geq0$.