How can we show $$1 + t^p \geq (1+t)^p$$
where $p < 1$ and $t>0$?
This inequality is useful to show $\sqrt{a+b} \leq \sqrt{a}+ \sqrt{b}$ for $a,b\geq 0$.
My try:
One can define $f(t)=1+t^p-(1+t)^p$ so $f'(t)=pt^{p-1}-p(1+t)^{p-1}$.
Therefore,
$$f'(t)=p(t^{p-1}-(1+t)^{p-1})=p(\frac{1}{t^{1-p}}-\frac{1}{(1+t)^{1-p}})$$
$$
t < 1 +t \rightarrow \frac{1}{t} > \frac{1}{1 +t}
$$
Now I am wondering what is the general rule for raising both sides of an equality to a positive number?
Is it true $|a|<|b| \rightarrow |a|^p<|b|^p \,\,\, \forall p>0$?
Can you help me to figure out the rule when both sides are not positive?
Best Answer
For $t=0$, equality holds. Now for $t>0$, look at the derivatives w.r.t. $t$ on both sides:
$$ (1+t^p)' = p t^{p-1} \\ ((1+t)^p)' = p (1+t)^{p-1} $$ As $0< p<1$, we have for all $t$ that $$ t < 1+ t \\ \log t < \log (1+t)\\ (p-1) \log t > (p-1) \log (1+t) \\ t^{p-1} > (1+t)^{p-1} \\ p t^{p-1} > p (1+t)^{p-1} \\ (1+t^p)' > ((1+t)^p)' $$
so, starting from equality at $t=0$, $1+t^p$ grows faster with $t$ than $(1+t)^p$ (for all $0<p<1$) which proves the claim.