## Is it possible to bring a rod of length \$ sqrt {3} l \$ into a side cube of length l?

A cube of side length l may be suitable for a rod of maximum length $$sqrt {3} l$$. But can such a rod be brought into a cube to be fixed in this position?

(Not even sure it's a mathematical question 🙂 and any help with the tags would be appreciated.)

## Prove \$ forall k geq 4, log (1 + x_k) -x_k leq {-1 over6k} \$ where \$ x_k = {(- 1) ^ k over sqrt k} \$

The question:

Prove $$forall k geq 4, log (1 + x_k) -x_k leq {-1 over6k}$$ or $$x_k = {(- 1) ^ k over sqrt k}$$.

The inequality above is valid if and only if begin {align} & log (1 + x_k) leq x_k- {1 on 6k} \ & Leftrightarrow 1 + x_k leq exp (x_k- {1 over6k}) end {align}
Using $$x + 1 leq e ^ x$$we have $$x_k + 1 leq e ^ {x_k}$$

## Calculate the maximum value of \$ sqrt { dfrac {3yz} {3yz + x}} + sqrt { dfrac {3zx} {3zx + 4y}} + sqrt { dfrac {xy} {xy + 3z}} where \$ x + 2y + 3z = 2 \$.

$$x$$, $$y$$ and $$z$$ are positive such as $$x + 2y + 3z = 2$$. Calculate the maximum value of $$large sqrt { dfrac {3yz} {3yz + x}} + sqrt { dfrac {3zx} {3zx + 4y}} + sqrt { dfrac {xy} {xy + 3z}}$$

This problem is brought to you by a recent contest. There should be different, more creative answers than the one I provided. Never mind…

## inequality – Calculates the minimum value of \$ sum_ {cyc} frac {a ^ 2} {b + c} \$ where \$ a, b, c> 0 \$ and \$ sum_ {cyc} sqrt {a ^ 2 + b ^ 2} = 1 \$.

$$a$$, $$b$$ and $$c$$ are positive such as $$sqrt {a ^ 2 + b ^ 2} + sqrt {b ^ 2 + c ^ 2} + sqrt {c ^ 2 + a ^ 2} = 1$$. Calculate the minimum value of $$large frac {a ^ 2} {b + c} + frac {b ^ 2} {c + a} + frac {c ^ 2} {a + b}$$

I do not have any ideas yet to solve the problem. I will probably do it in the near future, but for now, I can not.

## abstract algebra – Calculating the product of ideals \$ left (4, frac {3+ sqrt {-71}} {2} right) \$ and \$ left (3, frac {1+ sqrt {- 71 }} {2} right) \$

Let $$frak {a} = left (4, frac {3 + sqrt {-71}} {2} right)$$ and $$frak {b} = left (3, frac {1 + sqrt {-71}} {2} right)$$ to be the ideals in $$mathbb {Z}[sqrt{-71}]$$. I would like to show that $$frak {a} frak {b} = left (12, frac {-5 + sqrt {-71}} {2} right)$$. Clearly, $$frak {a} frak {b}$$ contains the ideal $$left (12, frac {-5 + sqrt {-71}} {2} right)$$. How about the conversation? Actually, I do not know how to write an arbitrary element $$frak {a} frak {b}$$ explicitly since $$mathbb {Z}[sqrt{-71}]$$ is not a PID. Thank you!

## algebra precalculus – If \$ x = frac { sqrt {111} -1} {2} \$, calculate \$ (2x ^ {5} + 2x ^ {4} – 53x ^ {3} – 57x + 54) ^ { 2004} \$.

I already have two solutions to this problem, it's for high school students with an advanced level. I would like to know if there are better or more creative approaches to the problem. Here are my solutions:

1. (1st solution): Note that
$$x ^ {2} = ( frac { sqrt {111} -1} {2}) ^ {2} = 28 – frac { sqrt {111}} {2}$$
$$x ^ {3} = x cdot x ^ {2} = left ( frac { sqrt {111}} {2} – frac {1} {2} right) left (28 – frac { sqrt {111}} {2} right) = 14 sqrt {111} – 111/4 – 14 + frac { sqrt {111}} {4} = frac {57 sqrt {111} } {4} – frac {167} {4}$$
$$x ^ {4} = (x ^ {2}) ^ {2} = frac {111 + 4 cdot 784} {4} – 28 sqrt {111}$$
$$x ^ {5} = x ^ {4} cdot x = left ( frac {111 + 4 cdot 784} {4} – 28 sqrt {111} right) left ( frac { sqrt {111} -1} {2} right) = left ( frac {3247} {4} – 28 sqrt {111} right) left ( frac { sqrt {111} -1} { 2} right)$$
$$= sqrt {111} frac {3359} {8} – frac {15679} {8}$$

So we have
$$2x ^ {5} + 2x ^ {4} – 53x ^ {3} – 57x + 54 =$$
$$( sqrt {111} frac {3359} {4} – frac {15679} {4}) + ( frac {111 + 4 cdot 784} {2} – 56 sqrt {111}) – 53 ( frac {57 sqrt {111}} {4} – frac {167} {4}) – 57 ( frac { sqrt {111} -1} {2} + 54$$
$$= sqrt {111} ( frac {3359} {4} – frac {3359} {4}) – frac {15679} {4} + frac {111 + 4 cdot 784} {2} + frac {53 cdot 167} {4} + frac {57} {2} + 54$$
$$= – frac {15679} {4} + frac {222 + 8 cdot 784} {4} + frac {53 cdot 167} {4} + frac {114} {4} + frac {216} {4}$$
$$= – frac {15679} {4} + frac {5600 + 894} {4} + frac {5300 + 3180 + 371} {4} + frac {114} {4} + frac {216} } {4}$$
$$= – frac {15679} {4} + frac {6494} {4} + frac {8851} {4} + frac {114} {4} + frac {216} {4}$$
$$= -4/4 = -1,$$
and the answer is $$(-1) ^ {2004} = 1.$$ The above solution requires tedious calculations. You will find below an alternative solution.

1. (2nd solution): Note that $$x = frac { sqrt {111} -1} {2}$$ is equivalent with
$$(2x + 1) ^ {2} = 111$$
$$4x ^ {2} + 4x + 1 = 111$$
$$4x ^ {2} + 4x – 110 = 0$$
$$(2x ^ {2} + 2x – 55) = 0 : : …….. : : (1)$$
Multiply $$(1)$$ with $$x ^ {3}$$ we have
$$(2x ^ {5} + 2x ^ {4} – 55x ^ {3}) = 0$$
Multiply $$(1)$$ with $$x$$ we have
$$(2x ^ {3} + 2x ^ {2} – 55x) = 0$$
Sum both and we get:
$$2x ^ {5} + 2x ^ {4} – 53 x ^ {3} + 2x ^ {2} – 55x = 0 : : …….. : : (2)$$
and now we have the first 3 terms of the form we want to calculate. Subtract $$(2)$$ with $$(1)$$ get:
$$2x ^ {5} + 2x ^ {4} – 53 x ^ {3} – 57x + 55 = 0$$
$$2x ^ {5} + 2x ^ {4} – 53 x ^ {3} – 57x + 54 = -1$$
So the answer is $$(- 1) ^ {2004} = 1.$$

## real analysis – \$ x_1 = sqrt {2} \$ and we have \$ x_ {n + 1} = ( sqrt {2}) ^ {x_n} \$. Discover the limit.

The limit of this sequence, if it exists, will be a fixed point of $$g (x) = sqrt {2} ^ x$$. So, if you solve the equation $$x = sqrt {2} ^ x$$you have all the possible limits (the solutions are 2 and 4). In the end, you have to use the fixed point theorem to establish convergence when you start with $$x_1 = sqrt {2}$$.

## Why is the \$ x ^ 2 \$ integral of \$ 0 to \$ 5 not equal to the \$ sqrt x \$ integral of \$ 0 to \$ 25?

My calculator says the first is 41.67 and the last is 83.34. Why is that? Should not they be equal?

## Show that: \$ displaystyle sum_ {cyc} frac { sqrt {x ^ 2 + y ^ 2}} {x ^ 2 + y ^ 2 + xy} ≥ sqrt {2} \$

CA watch:

$$displaystyle sum_ {cyc} frac { sqrt {x ^ 2 + y ^ 2}} {x ^ 2 + y ^ 2 + xy} ≥ sqrt {2}$$

Or $$a, b, c$$ in R \$

I'm trying to use the incumbent inequality

Please give me some ideas or tips

## Number Theory – Show that \$ sqrt {d} \$ has a length of period 1 ssi \$ d = a ^ 2 + 1 \$

Since I know if $$d$$ is an integer that $$sqrt {d} =[alpha_0,bar{alpha_1},…bar{alpha_n},bar{2alpha_0}]$$. I want to show that $$sqrt {d}$$ has a length of period 1 if and only if $$d = a ^ 2 + 1$$, for a natural number a.

This is what I think I should do. Could someone confirm it to me?

($$Leftarrow$$) Assumed $$d = a ^ 2 + 1$$then $$sqrt {d} = sqrt {a ^ 2 + 1}$$.

$$(a-1) ^ 2 leq a ^ 2 + 1, Rightarrow (a-1) leq sqrt {a ^ 2 + 1}$$

So the soil of $$sqrt {a ^ 2 + 1}$$ is $$a-1 = alpha_0$$.
By subtracting this and inverting, we obtain:

$$tfrac {1} { sqrt {a ^ 2 + 1} – (a-1)} = tfrac { sqrt {a ^ 2 + 1} + (a-1)} {2a}$$, which has floor 1.

So we know $$alpha_1 = 1$$, subtracting this and inverting, we obtain:

$$tfrac {2a} { sqrt {a ^ 2 + 1} – (a-1)} = tfac {2a ( sqrt {a ^ 2 + 1} + (a-1)} {2a} = sqrt {a ^ 2 + 1} + (a-1)$$who has the floor $$2a-2$$, subtracting and inverting gives,

$$tfrac {1} { sqrt {a ^ 2 + 1} – (a-1)$$So, we are back at the beginning and the continuous fraction has a period length of 1.

$$( Rightarrow)$$ assume $$sqrt {d}$$ for the length of period 1. Then $$sqrt {d} =[alpha_0,bar{alpha_1}]=[alpha_0,bar{alpha_1},bar{2alpha_0}]$$.

So we can write:

$$sqrt {d} = alpha_0 + tfrac {1} { alpha_1 + tfrac {1} {2 alpha_0 + tfrac {1} {…}}} = alpha_0 + tfrac {1} { alpha_1 + tfrac {1} {2 sqrt {d}}}$$.

I have not yet managed to solve this problem, but $$d = a ^ 2 + 1$$, do you come down to rearrange what I just wrote?