Math Genius: Does this $(x+y)^2ge 4xy$ hold for all real $x$ and $y$ – not only for non-negative?

Original Source Link

Does this $(x+y)^2ge 4xy$ hold for all real $x$ and $y$ – not only for non-negative?

I’m pretty much sure it is:

Suppose above is not true, that is $(x+y)^2< 4xy$

Then we come to a contradiction $(x-y)^2< 0$, which means that our assumption is false => the opposite is true.

Your reasoning is right.

I think it’s better the following.

Note $$(x+y)^2-4xy=(x-y)^2geq0$$
What do you think?

Your method is correct, but I think it can be simplified. In fact, you can rewrite the inequality as:
$$x^2+y^2+2xy-4xy>0leftrightarrow (x-y)^2>0$$
which is always true $forall x,y in R$.

Because ${left ( (x+ y)^{2}- 4xy right )}’= 2(x- y)(1- {y}’)$, so

$$(x+ y)^{2}- 4xygeq (y+ y)^{2}- 4y^{2}= 0$$

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Math Genius: Does this $(x+y)^2ge 4xy$ hold for all real $x$ and $y$ – not only for non-negative?

Original Source Link

Does this $(x+y)^2ge 4xy$ hold for all real $x$ and $y$ – not only for non-negative?

I’m pretty much sure it is:

Suppose above is not true, that is $(x+y)^2< 4xy$

Then we come to a contradiction $(x-y)^2< 0$, which means that our assumption is false => the opposite is true.

Your reasoning is right.

I think it’s better the following.

Note $$(x+y)^2-4xy=(x-y)^2geq0$$
What do you think?

Your method is correct, but I think it can be simplified. In fact, you can rewrite the inequality as:
$$x^2+y^2+2xy-4xy>0leftrightarrow (x-y)^2>0$$
which is always true $forall x,y in R$.

Because ${left ( (x+ y)^{2}- 4xy right )}’= 2(x- y)(1- {y}’)$, so

$$(x+ y)^{2}- 4xygeq (y+ y)^{2}- 4y^{2}= 0$$

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Math Genius: Proof of inequality $b^n-a^n<(b-a)nb^{n-1}$ when $0<a0$.

Original Source Link

I am working through some properties of $mathbb{R}$ and I stumbled upon the following theorem:

Theorem 1.21: For every real $x>0$ and every integer $n>0$, there is one and only one positive real $y$ such that $y^n=x$.

The proof of this theorem, as stated in Rudin’s Principles of Mathematical Analysis heavily relies on the following inequality:
$$
b^n-a^n < (b-a)nb^{n-1}, text{where} 0<a<b.
$$

I would really like to prove this inequality myself, and I tried rewriting the RHS. This produced
$$
b^n-a^n<n(b^n-acdot b^{n-1}).
$$

Clearly, the value given inside the parentheses on the RHS will be positive, as $b>a$, but I fail to see how this gives me the inequality itself. Does it maybe have to do something with the Archimedean Property of $mathbb{R}$?

Any help would be much appreciated.

note that b^n – a^n = (b-a)(b^n-1 + a*b^n-2 + a^2*b^n-3 + … + b*a^n-2 + a^n-1) …(1)
also since for any k>0 x^k is an increasing function , and hence a < b implies a^k < b^k
therefore , a^i * b ^(n-i-1) < or = b^i * b^(n-i-1) = b^n-1 for i = 0,1,2,…,n-1 …. (2).
So from (1) and (2) we have b^n – a^n < or =(b-a)nb^n-1 which is what we were required to prove.

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Math Genius: Proof of inequality $b^n-a^n<(b-a)nb^{n-1}$ when $0<a0$.

Original Source Link

I am working through some properties of $mathbb{R}$ and I stumbled upon the following theorem:

Theorem 1.21: For every real $x>0$ and every integer $n>0$, there is one and only one positive real $y$ such that $y^n=x$.

The proof of this theorem, as stated in Rudin’s Principles of Mathematical Analysis heavily relies on the following inequality:
$$
b^n-a^n < (b-a)nb^{n-1}, text{where} 0<a<b.
$$

I would really like to prove this inequality myself, and I tried rewriting the RHS. This produced
$$
b^n-a^n<n(b^n-acdot b^{n-1}).
$$

Clearly, the value given inside the parentheses on the RHS will be positive, as $b>a$, but I fail to see how this gives me the inequality itself. Does it maybe have to do something with the Archimedean Property of $mathbb{R}$?

Any help would be much appreciated.

note that b^n – a^n = (b-a)(b^n-1 + a*b^n-2 + a^2*b^n-3 + … + b*a^n-2 + a^n-1) …(1)
also since for any k>0 x^k is an increasing function , and hence a < b implies a^k < b^k
therefore , a^i * b ^(n-i-1) < or = b^i * b^(n-i-1) = b^n-1 for i = 0,1,2,…,n-1 …. (2).
So from (1) and (2) we have b^n – a^n < or =(b-a)nb^n-1 which is what we were required to prove.

Tagged : / /

Math Genius: Proof of inequality $b^n-a^n<(b-a)nb^{n-1}$ when $0<a0$.

Original Source Link

I am working through some properties of $mathbb{R}$ and I stumbled upon the following theorem:

Theorem 1.21: For every real $x>0$ and every integer $n>0$, there is one and only one positive real $y$ such that $y^n=x$.

The proof of this theorem, as stated in Rudin’s Principles of Mathematical Analysis heavily relies on the following inequality:
$$
b^n-a^n < (b-a)nb^{n-1}, text{where} 0<a<b.
$$

I would really like to prove this inequality myself, and I tried rewriting the RHS. This produced
$$
b^n-a^n<n(b^n-acdot b^{n-1}).
$$

Clearly, the value given inside the parentheses on the RHS will be positive, as $b>a$, but I fail to see how this gives me the inequality itself. Does it maybe have to do something with the Archimedean Property of $mathbb{R}$?

Any help would be much appreciated.

note that b^n – a^n = (b-a)(b^n-1 + a*b^n-2 + a^2*b^n-3 + … + b*a^n-2 + a^n-1) …(1)
also since for any k>0 x^k is an increasing function , and hence a < b implies a^k < b^k
therefore , a^i * b ^(n-i-1) < or = b^i * b^(n-i-1) = b^n-1 for i = 0,1,2,…,n-1 …. (2).
So from (1) and (2) we have b^n – a^n < or =(b-a)nb^n-1 which is what we were required to prove.

Tagged : / /

Math Genius: Does this $(x+y)^2ge 4xy$ hold for all real $x$ and $y$ – not only for non-negative?

Original Source Link

Does this $(x+y)^2ge 4xy$ hold for all real $x$ and $y$ – not only for non-negative?

I’m pretty much sure it is:

Suppose above is not true, that is $(x+y)^2< 4xy$

Then we come to a contradiction $(x-y)^2< 0$, which means that our assumption is false => the opposite is true.

Your reasoning is right.

I think it’s better the following.

Note $$(x+y)^2-4xy=(x-y)^2geq0$$
What do you think?

Your method is correct, but I think it can be simplified. In fact, you can rewrite the inequality as:
$$x^2+y^2+2xy-4xy>0leftrightarrow (x-y)^2>0$$
which is always true $forall x,y in R$.

Because ${left ( (x+ y)^{2}- 4xy right )}’= 2(x- y)(1- {y}’)$, so

$$(x+ y)^{2}- 4xygeq (y+ y)^{2}- 4y^{2}= 0$$

Tagged : / / / /

Math Genius: Does this $(x+y)^2ge 4xy$ hold for all real $x$ and $y$ – not only for non-negative?

Original Source Link

Does this $(x+y)^2ge 4xy$ hold for all real $x$ and $y$ – not only for non-negative?

I’m pretty much sure it is:

Suppose above is not true, that is $(x+y)^2< 4xy$

Then we come to a contradiction $(x-y)^2< 0$, which means that our assumption is false => the opposite is true.

Your reasoning is right.

I think it’s better the following.

Note $$(x+y)^2-4xy=(x-y)^2geq0$$
What do you think?

Your method is correct, but I think it can be simplified. In fact, you can rewrite the inequality as:
$$x^2+y^2+2xy-4xy>0leftrightarrow (x-y)^2>0$$
which is always true $forall x,y in R$.

Because ${left ( (x+ y)^{2}- 4xy right )}’= 2(x- y)(1- {y}’)$, so

$$(x+ y)^{2}- 4xygeq (y+ y)^{2}- 4y^{2}= 0$$

Tagged : / / / /

Math Genius: Finding a constant in inequality for softmax function

Original Source Link

Let $x, y$ be vectors in $mathbb{R}^n$, and $textrm{softmax}: mathbb{R}^n to mathbb{R}^n$ is a vector function such that:
$$textrm{softmax}(x)_i = frac{e^{x_i}}{sum^n_{j=1} e^{x_j}}.$$

What is the lowest constant $L$ such that:
$$forall x, y: |textrm{softmax} (x) – textrm{softmax}(y)|_{infty} leq L |x – y |_1 $$?

I have tried to invert the softmax function, but I am not sure what to do with the result or even if it is a useful thing to do.

I shall be grateful for any help in this task.

Edit: There is the property that

$$forall c in mathbb{R}: textrm{softmax}(x + c) = textrm{softmax}(x),$$

where $x + c = (x_1 + c, dots, x_n + c)$.

Then we need to find such $L$ that
$$|textrm{softmax}(x) – textrm{softmax}(y)|_{infty} leq L
|(x + c_x) – (y + c_y) |_1 = L
|x – y + c|_1$$

for any $x, y in mathbb{R}^n, c in mathbb{R}$.

The right-hand side is minimal when $c = textrm{median} left(x_1 – y_1, dots, x_n – y_n right)$.

I am not sure, but maybe this can be more convenient for derivation?

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Math Genius: Prove $(a+b+c)^3 (a+b-c)(b+c-a)(c+a-b) leqq 27a^2 b^ 2 c^2$

Original Source Link

For $a,b,c>0$$,$ prove$:$ $$(a+b+c)^3 (a+b-c)(b+c-a)(c+a-b) leqq 27a^2 b^ 2 c^2$$
My proof by S-S method$,$ see here.

Another proof by $pqr$ method$:$

Let $p=a+b+c,,q=ab+bc+ca,, r=abc.$ This inequality equivalent to$:$ $${p}^{6}-4,{p}^{4}q+8,{p}^{3}r+27,{r}^{2} geqq 0$$

Or$:$ $${frac { left( {p}^{4}-5,{p}^{2}q+6,pr+4,{q}^{2} right)
left( 7,{p}^{4}+45,{p}^{2}q+54,pr-36,{q}^{2} right) }{12{p}^{2}}}
+,{frac { left( {p}^{2}-3,q right) left( 5,{p}^{2}-3,q
right) left( {p}^{2}-4,q right) ^{2}}{12{p}^{2}}} geqq 0$$

Which is obvious because $p^2 geqq 3q,, p^4 -5p^2 q+6pr+4q^2 geqq 0 ,(text{Schur degree 4})$

I hope for another proof (without $uvw$!). Thanks for a real lot!

PS$:$ You can get $pqr$‘s form more faster by using Maple$,$ see here.

If $a+b-c<0$ and $a+c-b<0$, we obtain $a<0,$ which is a contradiction.

Thus, it’s enough to prove our inequality for $a+b-c>0$, $a+c-b>0$ and $b+c-a>0.$

Now, let $a+b-c=z$, $a+c-b=y$ and $b+c-a=x$.

Thus, we need to prove that $$27(x+y)^2(x+z)^2(y+z)^2geq64xyz(x+y+z)^3,$$ which follows from $$(x+y)(x+z)(y+z)geqfrac{8}{9}(x+y+z)(xy+xz+yz).$$
Can you end it now?

Geometric approach :

Let $a,b,c$ be the side of an triangle $ABC$ then your inequality is :
$$Tleqfrac{3sqrt{3}abc}{4(a+b+c)}$$
Or $$4sqrt{3}Tleq frac{9abc}{a+b+c}$$
Where $T$ is the area of the triangle $ABC$

For a proof see (found on Wikipedia):
Posamentier, Alfred S. and Lehmann, Ingmar. The Secrets of Triangles, Prometheus Books, 2012.

Suppose $a=max{a,b,c}.$

If $a geqslant b + c$ then
$$text{LHS} leqslant 0 leqslant text{RHS}.$$
If $a leqslant b + c,$ we write inequality as
$$ 27 cdot a(b+c-a)cdot b(c+a-b)cdot c(a+b-c) cdot (a+b+c)^3 leqslant 9^3 cdot a^3b^3c^3.$$
By the AM-GM inequality we have
$$27 cdot a(b+c-a)cdot b(c+a-b)cdot c(a+b-c) leqslant left[a(b+c-a)+b(c+a-b)+c(a+b-c)right]^3.$$
So, we need to prove
$$(a+b+c)left[2(ab+bc+ca)-a^2-b^2-c^2right] leqslant 9abc.$$
It’s Schur inequality.

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Math Genius: Prove $(a+b+c)^3 (a+b-c)(b+c-a)(c+a-b) leqq 27a^2 b^ 2 c^2$

Original Source Link

For $a,b,c>0$$,$ prove$:$ $$(a+b+c)^3 (a+b-c)(b+c-a)(c+a-b) leqq 27a^2 b^ 2 c^2$$
My proof by S-S method$,$ see here.

Another proof by $pqr$ method$:$

Let $p=a+b+c,,q=ab+bc+ca,, r=abc.$ This inequality equivalent to$:$ $${p}^{6}-4,{p}^{4}q+8,{p}^{3}r+27,{r}^{2} geqq 0$$

Or$:$ $${frac { left( {p}^{4}-5,{p}^{2}q+6,pr+4,{q}^{2} right)
left( 7,{p}^{4}+45,{p}^{2}q+54,pr-36,{q}^{2} right) }{12{p}^{2}}}
+,{frac { left( {p}^{2}-3,q right) left( 5,{p}^{2}-3,q
right) left( {p}^{2}-4,q right) ^{2}}{12{p}^{2}}} geqq 0$$

Which is obvious because $p^2 geqq 3q,, p^4 -5p^2 q+6pr+4q^2 geqq 0 ,(text{Schur degree 4})$

I hope for another proof (without $uvw$!). Thanks for a real lot!

PS$:$ You can get $pqr$‘s form more faster by using Maple$,$ see here.

If $a+b-c<0$ and $a+c-b<0$, we obtain $a<0,$ which is a contradiction.

Thus, it’s enough to prove our inequality for $a+b-c>0$, $a+c-b>0$ and $b+c-a>0.$

Now, let $a+b-c=z$, $a+c-b=y$ and $b+c-a=x$.

Thus, we need to prove that $$27(x+y)^2(x+z)^2(y+z)^2geq64xyz(x+y+z)^3,$$ which follows from $$(x+y)(x+z)(y+z)geqfrac{8}{9}(x+y+z)(xy+xz+yz).$$
Can you end it now?

Geometric approach :

Let $a,b,c$ be the side of an triangle $ABC$ then your inequality is :
$$Tleqfrac{3sqrt{3}abc}{4(a+b+c)}$$
Or $$4sqrt{3}Tleq frac{9abc}{a+b+c}$$
Where $T$ is the area of the triangle $ABC$

For a proof see (found on Wikipedia):
Posamentier, Alfred S. and Lehmann, Ingmar. The Secrets of Triangles, Prometheus Books, 2012.

Suppose $a=max{a,b,c}.$

If $a geqslant b + c$ then
$$text{LHS} leqslant 0 leqslant text{RHS}.$$
If $a leqslant b + c,$ we write inequality as
$$ 27 cdot a(b+c-a)cdot b(c+a-b)cdot c(a+b-c) cdot (a+b+c)^3 leqslant 9^3 cdot a^3b^3c^3.$$
By the AM-GM inequality we have
$$27 cdot a(b+c-a)cdot b(c+a-b)cdot c(a+b-c) leqslant left[a(b+c-a)+b(c+a-b)+c(a+b-c)right]^3.$$
So, we need to prove
$$(a+b+c)left[2(ab+bc+ca)-a^2-b^2-c^2right] leqslant 9abc.$$
It’s Schur inequality.

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