Is it possible to prove the following inequality?

Question

I have the following inequality $$\frac{(1+x(c-1))^2}{(1+x^{\frac{2}{m}}(c-1))^m}\geq c^{2-m}$$ where $0<x<1$, $c$ is some positive integer $\geq 1$ and $m>2$. I tried to plot the function for different values and it seems that the inequality is true however I do not know whether it is true for all possible values of the parameters. Any help in this regard will be much appreciated. Thanks in advance.

Answer 1

your inequality is equivalent to $$\left( \frac 1c+x\left( 1-\frac 1c \right) \right)^2\geqslant \left( \frac 1c+x^{2/m}\left( 1-\frac 1c \right) \right)^m,$$ note that the power mean inequality give: $\forall x$, $y>0$, $0\leqslant \lambda \leqslant 1$, $p\geqslant q$, $pq\ne0$, then $$\bigl( \lambda x^p+(1-\lambda )y^p \bigr)^{1/p}\geqslant \bigl( \lambda x^q+(1-\lambda )y^q \bigr)^{1/q},$$ so, your inequality holds for $x\in(0,+\infty)$, $c\in[1,+\infty)$, $m\in(-\infty,0)\cup[2,+\infty)$.

Answer 2

I will provide the proof of the general inequality which kuing used in the answer. This is how I was initially trying to prove the OP case.

Start with:

$$\left( \lambda x^p+(1-\lambda )y^p \right)^{1/p} \geq \left( \lambda x^q+(1-\lambda )y^q \right)^{1/q}$$

The conditions on the parameters are more restricted, namely $x>y>0$, $\lambda \in (0,1)$ and $p>q$ (the equality cases are trivial, so I leave them out).

First, we simplify the expression by introducing $t=y/x \in (0,1)$, then taking logarithms (using convexity of logarithm function) and considering:

$$f(t)=\frac{1}{p} \ln \left( \lambda +(1-\lambda )t^p \right)-\frac{1}{q} \ln \left( \lambda +(1-\lambda )t^q \right)$$

Second, we check if there are any local extrema for $f(t)$ on the interval $t \in (0,1)$:

$$f'(t)=\frac{1-\lambda}{t} \left( \frac{t^p}{\lambda +(1-\lambda )t^p}-\frac{t^q}{\lambda +(1-\lambda )t^q} \right)$$

$$f'(t)=\frac{(1-\lambda) \lambda}{t(\lambda +(1-\lambda )t^p)(\lambda +(1-\lambda )t^q)} \left( t^p-t^q \right) < 0$$

$$(p>q)$$

Since there are no local extrema on the whole interval, the minimal/maximal values can only be achieved at endpoints. Since the derivative is negative, the function is decreasing on the interval, so its minimum will be achieved at $t \to 1$.

Let's check this endpoint:

$$t=1+\epsilon$$

$$f(t)=\frac{1}{p} \ln \left( \lambda +(1-\lambda )(1+\epsilon)^p \right)-\frac{1}{q} \ln \left( \lambda +(1-\lambda )(1+\epsilon)^q \right)$$

$$\lim_{\epsilon \to 0} f(t)=0$$

Which means that:

$$f(t) >0 , \qquad t \in (0,1)$$

Which proves the original inequality.