# How to prove $a x^{2a} -2 a x^a \geq x^{a+1}-(a+1)x$

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?

Thanks

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 $$(0,1]$$.

• Thanks. But my question is exactly how to study? Nov 11, 2020 at 9:06
• @user295959: I have expanded my hint
– 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 $$0 such that $$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.