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?
 A: Try to show that $(1+y^4)\le (1+|y|)^4$ insteadd
A: Consider the quadratic $(y+ \frac{3}{4})^2+ \frac{7}{16}$, which has a minimum value at $(-3/4, 7/16)$. Thus $$(y+ \frac{3}{4})^2+ \frac{7}{16} \geq 0$$
For $y>0$, $$4y\left((y+ \frac{3}{4})^2+ \frac{7}{16}\right) \geq 0$$ is also true.
Expanding the brackets, we are left with
$$4y^3+6y^2+4y \geq 0 $$
Which is equivalent to
$$1+y^4 \leq 1+4y+6y^2+4y^3+y^4$$
$$1+y^4 \leq (1+y)^4$$
$$ \sqrt[4]{1+y^4} \leq 1+y$$
For $y>0$, $y = |y|$
So the inequality is true for $y>0$.
Considering the original inequality,
$$ \sqrt[4]{1+y^4} \leq 1+|y|$$
we can see that both functions of $y$, $y^4$ and $|y|$ are even (i.e. The same for both positive and negative values of y), thus, if the inequality is true for $y>0$, it is true for $y<0$ as well.
The only remaining case is for $y=0$, which can be verified simply by simply plugging $y=0$ into the original inequality.
(I hope that this is okay as an answer, this is my first post on Stack Exchange so I hope I have followed the guidelines!)
A: 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
$.
