Question

How would you prove, using Newton's binomial theorem, that
\[\forall (x,y) \in \mathbb{R}_{+} \times \mathbb{R}{+}, \sqrt[n]{x+y}\le \sqrt[n]{x}+\sqrt[n]{y}\]

Answer 1

u got 2 more clear

Answer 2

how?

Answer 3

Would a substitution like
\[a=\frac{1}{n}\]
work and then just working with proving that
\[(x+y)^a \le x^a+y^a \]
Would that be equivalent?

Answer 4

hmm..i dont think that would work. expanding \[(x+y)^a\] as an integer would be different than expanding it as \[(x+y)^\frac{1}{n}\]

Answer 5

hmm... I have no clue, that was my idea hehe but coming to think of it the binomial formula works for an exponent that's an integer...

Answer 6

right. im reading some random webpages online, and it seems like there is a way to use the binomial theorem with rational exponents. I have no idea how to calculate the coefficients though =/

Answer 7

Oh, well... How would raising both sides to the nth work?
Say, having thus to prove that
\[x+y \le (\sqrt[n]{x}+\sqrt[n]{y})^n\]
And THEN using the binomial theorem to expand the expression with the roots... I think you'd get something like x+y+a lot of coefficients, no? Then the result would be proven! Is my logic correct?

Answer 8

that sounds like a good plan :)

Answer 9

thanks!

Answer 10

this looks like a fun problem but i have to go to bed :(

Answer 11

i have a paper to type for art history >.> i need to get off this site!

Answer 12

its an addiction its hard

Answer 13

lol >.< well, im going to log off for once, maybe that will help (i know it wont lol).

Answer 14

hey someone if someone doesn't give you an answer i will try to think about it tomorrow

Answer 15

i cant promise i will be able to do it but i will certainly try! :)

Answer 16

Myinin is the nest.

Answer 17

best*

Answer 18

is n a positive integer?

Answer 19

Yep, forgot to say n is a natural number, although being named n I think it's implied :p
Here I go for what I thought as proof:

Let f be the function
\[f(x)=x^n\]
f is an increaasing function in R+ which allows us to say
\[\sqrt[n]{x+y}\le \sqrt[n]{x}+\sqrt[n]{y} \Leftrightarrow x+y \le ( \sqrt[n]{x}+\sqrt[n]{y})^n\]
But
\[(\sqrt[n]{x}+\sqrt[n]{y})^n=\sum_{k=0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) \sqrt[n]x^{n-k} \sqrt[n]y^k=x+y + \sum_{k=1}^{n-1}\left(\begin{matrix}n \\ k\end{matrix}\right) \sqrt[n]x^{n-k} \sqrt[n]y^k\]

And knowing that
\[\forall (x,y) \in \mathbb{R}_+ \times \mathbb{R}_+ ,\sum_{k=1}^{n-1}\left(\begin{matrix}n \\ k\end{matrix}\right) \sqrt[n]x^{n-k} \sqrt[n]y^k \ge 0\]

Thus
\[x+y \le (\sqrt[n]{x} + \sqrt[n]{x})^n \Leftrightarrow x+y \le x+y + \sum_{k=1}^{n-1}\left(\begin{matrix}n \\ k\end{matrix}\right) \sqrt[n]x^{n-k} \sqrt[n]y^k\]
Substracting x+y from both sides gives
\[0 \le \sum_{k=1}^{n-1}\left(\begin{matrix}n \\ k\end{matrix}\right) \sqrt[n]x^{n-k} \sqrt[n]y^k\]
Which is true, meaning the initial statement is also true. Is this correct?

Answer 20

The Idea you have here is pretty much what I have on paper, but I didn't write it out the way you did....it does not flow nicely.

I would say something like since

\[\forall (x,y) \in \mathbb{R}_+ \times \mathbb{R}_+ ,\sum_{k=1}^{n-1}\left(\begin{matrix}n \\ k\end{matrix}\right) \sqrt[n]x^{n-k} \sqrt[n]y^k \ge 0\]

we have

\[x+y \le x+y + \sum_{k=1}^{n-1}\left(\begin{matrix}n \\ k\end{matrix}\right) \sqrt[n]x^{n-k} \sqrt[n]y^k=\sum_{k=0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) \sqrt[n]x^{n-k} \sqrt[n]y^k=(\sqrt[n]{x}+\sqrt[n]{y})^n\]

Answer 21

I did say that, although I admit my way to write it was pretty lazy hahaha, I think yours is cleaner!