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?
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1why not taking fourth power on both sides? – user729 Oct 20 '14 at 19:29
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I actually solved this moments after posting the question and forgot to update. I feel so remarkably stupid for asking this now.. Thank you for your comment though! – tachyon_man Oct 20 '14 at 19:36
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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!)
WJG
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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 $.
marty cohen
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