How to show that $x^{\tfrac{1}{p}}\leq \frac{1}{p}x+\frac{1}{q}$

Question

Let $p,q>1$, $\frac{1}{p}+\frac{1}{q}=1$ and $x\in[0,1]$. I would like to show that $x^{\tfrac{1}{p}}\leq \frac{1}{p}x+\frac{1}{q}$.

I tried to rewrite both sides of the inequality just involving $p$ or just involving $q$ but that didn't help me. Any hints how to prove the inequality?

Answer 1

Hint: Let $$f(x)=\frac{1}{p}x+\frac{1}{q}-x^{1/p}$$ then $$f'(x)=\frac{1}{p}-\frac{1}{p}x^{\frac{1}{p}-1}$$

Answer 2

We have that

$$ x^{\tfrac{1}{p}}\le \frac{1}{p}x+\frac{1}{q}\iff \frac1p \log x +\frac1q \log 1 \le \log \left(\frac1p x+\frac1q \right)$$

then recall that $\log$ function is concave and refer to Jensen's inequality .

Answer 3

Bernoulli's inequality: $(1+x)^r \leqslant 1+rx$ for $x \geqslant -1$ and $r \in [0,1]$.

Therefore: $~\displaystyle x^{\tfrac{1}{p}} \leqslant 1+\frac{1}{p}(x-1) = \frac{1}{p}+\frac{1}{q}+\frac{1}{p}(x-1) = \frac{1}{p}x+\frac{1}{q}.$

Answer 4

By the weighted AM-GM inequality with $\lambda_1 = \frac{1}{p}$, $x_1 = x$, $\lambda_2 = \frac{1}{q}$, $x_2 = 1$, we have $$x^{1/p} \cdot 1^{1/q} \le \frac{1}{p} \cdot x + \frac{1}{q} \cdot 1.$$