For $x \in [0, 1]$ and $a \geq 1$, how to prove $$a x^{2a} -2 a x^a \geq x^{a+1}-(a+1)x?$$

Because the inequality holds with equality at $x = 0$ and $x = 1$, we can not show that for each $a$, the function LHS - RHS is increasing / decreasing in $x$.

Because the inequality holds with equality at $a = 1$ for any $x$, this suggests that we can show, for each $x$, the LHS - RHS is increasing in $a$. But it is not easy to show this because the derivative is messy.

Is there any easy, elegant way to show this inequality?



2 Answers 2


HINT: You plugged $x=0$ and checked it works. So exclude this value and work on $(0,1]$; in this way you can divide by $x$ to get something easier to handle; at this point, study the sign of $$ f(x)=ax^{2a-1}-2ax^{a-1}-x^{a}+(a+1)\;. $$ taking the derivative: $$ f'(x)=a(2a-1)x^{2(a-1)}-2ax^{a-2}-x^{a-1} =x^{a-1}\underbrace{\left[a(2a-1)x^{a-1}-2ax^{-1}-1\right]}_{=:g(x)} $$ and since $x^{a-1}$ is always positive, you can just study the sign of the functions I have named $g$: \begin{align*} g(x)\ge0 &\Longleftrightarrow a(2a-1)x^{a-1}\ge2ax^{-1}+1\\ &\Longleftrightarrow a(2a-1)x^{a}\ge2a+x\\ \end{align*} where in the last step I have multiplied both sides by $x$ since it is strictly positive on $(0,1]$.

  • $\begingroup$ Thanks. But my question is exactly how to study? $\endgroup$
    – user295959
    Nov 11, 2020 at 9:06
  • $\begingroup$ @user295959: I have expanded my hint $\endgroup$
    – Joe
    Nov 11, 2020 at 9:17

The main idea is correct. Actually, the derivative is not messy with careful factoring.

Suppose $a$ is fixed. Define $$f(x)=ax^{2a}-2ax^a-x^{a+1}+(a+1)x$$ Since you have found that $f(1)=0$ and $f(0)=0$, we may think we can do some thing with its monoticity! If we can prove $f(x)$ increases at first and decreases to $0$ when $x=1$, $f(x)\ge 0$ will hold.

Thus its derivative can help us a lot: \begin{align} f'(x)&=2a^2x^{2a-1}-2a^2x^{a-1}-(a+1)x^a+a+1\\ &=2a^2x^{a-1}(x^a-1)-(a+1)(x^a-1)\\ &=(2a^2x^{a-1}-a-1)(x^a-1) \end{align} Define two new function $h(x)=2a^2x^{a-1}-a-1$ and $g(x)=x^a-1$, and then see how their value goes on $[0,1]$

Of course, $g(x)\le 0$ on $[0,1]$. While as to $h(x)$, it is an increasing fuction on $[0,1]$ and $h(0)=-a-1<0$, $h(1)=2a^2-a-1=(2a+1)(a-1)\ge0$. In this way, we should be able to find an $0<x_0<1$ such that $h(x_0)=0$. Therefore $h(x)\ge0$ on $[0,x_0)$ and $h(x)\le0$ on $(x_0,1]$.

Since $f'(x)=g(x)h(x)$, we can get that $f'(x)\ge0$ on $[0,x_0)$ and $f'(x)\le0$ on $(x_0,1]$. Thus, $f(x)$ will increase on $[0,x_0)$ and decrease on $(x_0,1]$, which fits the assumption we made at the beginning!

Factoring can help a lot in this kind of questions.


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