If $0<x<1$ then prove that $x^a \leq x < 1$ for all $a\in \mathbb{R}, a\geq 1$. If this is true then can somebody please help me to get the proof. Thanks!
I am trying to see the proof of this for all $a\in \mathbb{R}, a\geq 1$.I saw that this is easy inequality to prove if $a\in \mathbb{N}, a\geq 1$.  It follows easily by induction.
 I was trying to do this for $a\in \mathbb{R}, a\geq 1$. Now, I have figured out the proof which I have given below in this thread.
I have a proof for this inequality for all $a \geq 1, a\in \mathbb{R}$. Fix $0&lt;x&lt;1$ and 
define a function $f(a) = x^a - x$. Then $f^{\prime}(a) = x^a \ln x-0&lt;0$. Note that $\ln x &lt;0$ for $0&lt;x&lt;1$. So the function is decreasing for $a \geq 1$. That is $f(a) \leq f(1)$ as desired. 
This requires no induction and more powerful than induction.
If $a &gt; 0$ and $b &lt; c$, then $ab &lt; ac$.

So, for any integer $n\geq 1$: if $x^{n-1} &lt; x^n$, we must also have
$$
x\cdot x^{n-1} &lt; x \cdot x^{n}
$$
which leads (by induction) to the desired result.