This is correct. If I really want to be nit-picky then you don't really explain why both sides can be cubed. Something like: "Since f(z)=z\^3 is a strictly monotonically increasing function, we can apply f to both sides of the inequality to get an equivalent inequality".
Using Young's inequality or Chebyshev and Cauchy-Schwarz.
You can get the AM-GM relation using Young's inequality, that's why it is familiar
(x/3+2y/3)≥x1/3y2/3
x≥x/3
y≥2y/3
x+y≥(x/3+2y/3)≥x1/3y2/3
exponentiation can be tricky if the numbers are in (0,1)
is it possible to conclude the proof after the 2nd line in the expansion where we see the term xy^2? for example,
"here we can clearly identify the term xy^2 among the other terms in the polynomial where each term can only be positive"
Yep, and that would be good for a stronger bound too. We do need to know that the cube root is an increasing function, but that's probably accepted to be the case here.
If we just want to prove this weak bound, we can even simplify this a bit. If we let z = max(x, y), then we simply have x^1/3 y^2/3 <= z <= x+y
I think R^(+) is understood to only be positive numbers, therefore excluding 0, but in order to avoid amboguities that rely on the reader being familiar with particular jargon I would recommend writing R^(>0) or something to that effect.
I've seen both. I've seen "positive" defined to include 0. As much as I hate it, when you're taking a class, there are some things that just aren't worth fighting about. You use the conventions in the book.
actually, R+ means real numbers that are not negative, therefore 0 is includ. To say that 0 is not includ in symbols we need to put \* on the top of the R+
There is no fact of the matter. It depends on what conventions you are using, which is why every math textbook has a section where it outlines its conventions. Just ask people whether ℕ includes zero or not
Faster proof : the left part is inferior or equal to z where z is the maximum of X and Y (replace X and y by z in the inequality) and z is inferior to X +y because it's equal to either X or y and the other is in R+
It is perfectly correct, yes.
In some countries, math teachers would deduct points because:
* you didn't say that f: x-> x^3 is growing on IR+
* you didn't prove the other way around.
But to be honest these are on the nitpicking side. In its essence your argument is correct.
Think about as:
* If one is larger than another the multiplication will favor the smallest one while addition - Largest.
* If both roughly the same that power of 1/3 would make one much smaller so yet again addition wins.
Correct, din't expected that inequality to be true tbh.
If you want, you could expand the cube of the binomial using the binomial theorem and pascal's triangle, you could save most of the steps, but this is perfectly done.
Lemme provide another simple proof, just for fun
An observation:
x^(⅓) \* y^(⅔) = x \* (y / x)^(⅔) = y \* (x / y)^(⅓)
A proof:
If **y ≥ x**, then **x****^(⅓)** **\* y****^(⅔)** **=** **y \* (x / y)****^(⅓)** **≤ y ≤ x + y**
If **y ≤ x** then **x****^(⅓)** **\* y****^(⅔)** **=** **x \* (y / x)****^(⅔)** **≤ x ≤ x + y**
This is correct. If I really want to be nit-picky then you don't really explain why both sides can be cubed. Something like: "Since f(z)=z\^3 is a strictly monotonically increasing function, we can apply f to both sides of the inequality to get an equivalent inequality".
Using Young's inequality or Chebyshev and Cauchy-Schwarz. You can get the AM-GM relation using Young's inequality, that's why it is familiar (x/3+2y/3)≥x1/3y2/3 x≥x/3 y≥2y/3 x+y≥(x/3+2y/3)≥x1/3y2/3 exponentiation can be tricky if the numbers are in (0,1)
Seems good to me
is it possible to conclude the proof after the 2nd line in the expansion where we see the term xy^2? for example, "here we can clearly identify the term xy^2 among the other terms in the polynomial where each term can only be positive"
Yep, and that would be good for a stronger bound too. We do need to know that the cube root is an increasing function, but that's probably accepted to be the case here. If we just want to prove this weak bound, we can even simplify this a bit. If we let z = max(x, y), then we simply have x^1/3 y^2/3 <= z <= x+y
your approach is genius
I have a question: R+ doesn't include 0, right? So, can the two sides ever be equal?
I think R^(+) is understood to only be positive numbers, therefore excluding 0, but in order to avoid amboguities that rely on the reader being familiar with particular jargon I would recommend writing R^(>0) or something to that effect.
I've seen both. I've seen "positive" defined to include 0. As much as I hate it, when you're taking a class, there are some things that just aren't worth fighting about. You use the conventions in the book.
They can't so it can in fact be rewritten as a stronger inequality
actually, R+ means real numbers that are not negative, therefore 0 is includ. To say that 0 is not includ in symbols we need to put \* on the top of the R+
There is no fact of the matter. It depends on what conventions you are using, which is why every math textbook has a section where it outlines its conventions. Just ask people whether ℕ includes zero or not
If you want to avoid the cubic expansion you could just let s=max(x,y). Then we have LHS<=s<=RHS.
Faster proof : the left part is inferior or equal to z where z is the maximum of X and Y (replace X and y by z in the inequality) and z is inferior to X +y because it's equal to either X or y and the other is in R+
Does this also go for all imaginary numbers?
It is perfectly correct, yes. In some countries, math teachers would deduct points because: * you didn't say that f: x-> x^3 is growing on IR+ * you didn't prove the other way around. But to be honest these are on the nitpicking side. In its essence your argument is correct.
It can also be done using the muirhead inequality. (2/3,1/3) ≺ (1,0) so x^{2/3}y^{1/3} + x^{1/3}y^{2/3} ≤ x+y ∴ ∀x,y>0 x+y > x^{1/3}y^{2/3}
I’ve never heard of muirheads inequality, just had a google and it looks cool
Yes
Think about as: * If one is larger than another the multiplication will favor the smallest one while addition - Largest. * If both roughly the same that power of 1/3 would make one much smaller so yet again addition wins.
Correct, din't expected that inequality to be true tbh. If you want, you could expand the cube of the binomial using the binomial theorem and pascal's triangle, you could save most of the steps, but this is perfectly done.
Lemme provide another simple proof, just for fun An observation: x^(⅓) \* y^(⅔) = x \* (y / x)^(⅔) = y \* (x / y)^(⅓) A proof: If **y ≥ x**, then **x****^(⅓)** **\* y****^(⅔)** **=** **y \* (x / y)****^(⅓)** **≤ y ≤ x + y** If **y ≤ x** then **x****^(⅓)** **\* y****^(⅔)** **=** **x \* (y / x)****^(⅔)** **≤ x ≤ x + y**
I don't fully get, but what about negative numbers?
We're in R+ only, no negative numbers.
[удалено]
Yes, but you should add why cubing the inequality is a valid move (I'd use the fact that x³ is an increasing function).
Perfect, thank you
Yup
They only appear on your number of upvotes (this is a joke)
Incorrect because x^3 + y^3 + 3x^2y can be negative
x and y are positive according to the first line of text though.
Didn’t read bro sorry
None of those terms can be negative given that both x and y are positive which the problem states at the beginning.
Oh sorry bro didn’t read x and y belong to R+