MHB Can you prove this inequality challenge?

anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Prove that $\sqrt[n]{1+\dfrac{\sqrt[n]{n}}{n}}+\sqrt[n]{1-\dfrac{\sqrt[n]{n}}{n}}<2$ for any positive integer $n>1$.
 
Mathematics news on Phys.org
First note, that the function $f(x) = x^{\frac{1}{n}}$ is concave (downward) for $n = 2,3,4 …$.

Applying Jensens inequality for a concave function:

\[\frac{1}{N}\sum_{i=1}^{N}f(x_i)\leq f\left ( \frac{\sum_{i=1}^{N}x_i}{N} \right )\]

Equality holds if and only if $x_1 = x_2 = … = x_N$ or $f$ is linear.

In our case: $N = 2$ and $1+\frac{\sqrt[n]{n}}{n} \neq 1-\frac{\sqrt[n]{n}}{n}$ for all positive integers $n$.

Note also, that the case $n = 1$ implies, that $f$ is linear, which is why the case is omitted. Hence, we obtain:

\[\sqrt[n]{1+\frac{\sqrt[n]{n}}{n}} + \sqrt[n]{1-\frac{\sqrt[n]{n}}{n}} < 2\sqrt[n]{\frac{1+\frac{\sqrt[n]{n}}{n}+1-\frac{\sqrt[n]{n}}{n}}{2}} = 2\] q.e.d.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Thread 'Imaginary Pythagorus'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top