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

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.

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

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

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

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.

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

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

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

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

I would really like to prove this inequality myself, and I tried rewriting the RHS. This produced
$$b^n-a^n
Clearly, the value given inside the parentheses on the RHS will be positive, as $$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.

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

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

I would really like to prove this inequality myself, and I tried rewriting the RHS. This produced
$$b^n-a^n
Clearly, the value given inside the parentheses on the RHS will be positive, as $$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.

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

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

I would really like to prove this inequality myself, and I tried rewriting the RHS. This produced
$$b^n-a^n
Clearly, the value given inside the parentheses on the RHS will be positive, as $$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.

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

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.

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

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

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

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.

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

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

## Math Genius: Finding a constant in inequality for softmax function

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?

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

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.

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

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.