Show $1+x^a \leq (1+x)^a$ for $a>1$ and $x > 0$

Question

Show that $$1+x^a \leq (1+x)^a$$ for $a>1$ and $x > 0$

I'm trying to solve the inequality $$1+x^a \leq (1+x)^a,$$ where $a$ is a real number such that $a>1$ and for $x > 0$

Answer 1

$x(a-1)\geq1$ is wrong for $a>1$ and $x>0$. Try $a=\frac{3}{2}$ and $x=1$.

By the way, your inequality is true.

Let $f(x)=(1+x)^a-x^a-1$.

Thus, $f'(x)=a(1+x)^{a-1}-ax^{a-1}=ax^{a-1}\left(\left(1+\frac{1}{x}\right)^{a-1}-1\right)>0$,

which gives $f(x)>f(0)=0$.

Done!

Answer 2

HINT:

$$\left(\frac{1}{1+x}\right)^a + \left( \frac{x}{1+x}\right)^a < \frac{1}{1+x} + \frac{x}{1+x}$$