Prove that $\sqrt[4]{1+y^4} \leq 1+|y|$ Prove that $\sqrt[4]{1+y^4} \leq 1+|y|$ for all real values of $y$. I attempted to show this by finding the power series expansion of $\sqrt[4]{1+y^4} $ and then relating that to $1+|y|$; however, I have made little progress. Any advise?

First,
for $z > 0$,
$\sqrt{1+z^2}
< 1+z
$
(by squaring both sides).
Putting
$z^2$ for $z$,
$\sqrt{1+z^4}
< 1+z^2
$.
Taking the
square root of both sides,
$\sqrt[4]{1+z^4}
< \sqrt{1+z^2}
$.
Combining these two
(but you can do that yourself).
By induction
$\sqrt[2^n]{1+z^{2^n}}
< \sqrt[2^{n-1}]{1+z^{2^{n-1}}}
...
< 1+z
$.
