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[733] Csimby

2012-05-22 18:43:47

Na szóval, legyenek a₁,a₂...,a_n poz. valósak, továbbá n $\geq$ 3 Biz.be:

$\sum_{i=1}^n \frac{a_1a_2\dots a_n}{na_i}\leq \left( \prod_{1\leq i < j \leq n} \frac{a_i+a_j}{2}\right)^{2/(n^2-n)},$

egyenlőség pontosan akkor ha a₁=...=a_n.

Előzmény: [730] Csimby, 2012-05-22 18:12:38

[732] Csimby	2012-05-22 18:33:21
Bocs, a bal oldalt szumma van!!
Előzmény: [731] Róbert Gida, 2012-05-22 18:23:32

[731] Róbert Gida	2012-05-22 18:23:32
n=2;a₁=a₂=8-re nem teljesül az egyenlőtlenség.
Előzmény: [730] Csimby, 2012-05-22 18:12:38

[730] Csimby

2012-05-22 18:12:38

Legyenek a₁,a₂,...,a_n pozitív valós számok. Biz. be:

$\prod_{i=1}^n \frac{a_1a_2\dots a_n}{na_i}\leq \left(\prod_{1\leq i < j \leq n} \frac{a_i+a_j}{2}\right)^{2/(n^2-n)}$

Egyenlőség pontosan akkor ha a₁=a₂=...=a_n.

[729] sakkmath

2012-05-19 16:10:25

Az utoljára felvetett kérdésben csak idáig jutottam:

cot (1°).cot (2°).cot (3°).....cot (43°).cot (44°)=tan (46°).tan (47°).tan (48°).....tan (87°).tan (89°)

Ha valaki megkapta volna a szorzat számértékét, kérem, ne tartsa titokban :-)

Előzmény: [727] juantheron, 2012-05-18 17:02:03

[728] sakkmath	2012-05-19 00:25:39
This nostalgia for me: About 30 years ago, I proposed to publish these two problems. (Not published them ...)

Előzmény: [724] sakkmath, 2012-05-18 15:35:51

[727] juantheron

2012-05-18 17:02:03

Thanks Sakkmath i have got my solution after seeing your Answer

bcz I have solve a simillar Problem Yesterday where I have calculate value of sin (1⁰).sin (2⁰).sin (3⁰)......sin (89⁰)

Now I have another doubt in my min

can we calculate the value of

cot (1⁰).cot (2⁰).cot (3⁰)........cot (44⁰)=

[726] juantheron

2012-05-18 16:58:09

Let A=cos (1⁰).cos (2⁰).cos (3⁰).........cos (89⁰)

Now using cos (90-a)=sin (a)

So A=sin (1⁰).sin (2⁰).sin (3⁰).........sin (89⁰)

and Let B=sin (2⁰).sin (4⁰).sin (6⁰).........sin (88⁰)

Now AB=(sin (1⁰).sin (89⁰)).(sin (2⁰).sin (88⁰)).........(sin (44⁰).sin (46⁰)).sin (45⁰)

So AB=(sin (1⁰).cos (1⁰)).(sin (2⁰).cos (2⁰)).........(sin (44⁰).cos (44⁰)).sin (45⁰)

So $2^{44}AB=\sin (2^0).\sin(4^0)........\sin(88^0).\frac{1}{\sqrt{2}}$

So $2^{44}AB = \frac{1}{\sqrt{2}}B$

So $A=\frac{1}{\sqrt{2}.2^{44}}=\frac{\sqrt{2}}{2^{45}}$

bcz B $\neq$ 0

[725] juantheron

2012-05-18 16:57:24

Thanks sakkmath after seeing your answer I have got a method

bcz i have solve a simillar problem Yesterday where i have

calculate the value of

sin (1⁰).sin (2⁰).sin (3⁰).........sin (89⁰)

My solution::

Let A=cos (1⁰).cos (2⁰).cos (3⁰).........cos (89⁰)

Now using cos (90-a)=sin (a)

So A=sin (1⁰).sin (2⁰).sin (3⁰).........sin (89⁰)

and Let B=sin (2⁰).sin (4⁰).sin (6⁰).........sin (88⁰)

Now AB=(sin (1⁰).sin (89⁰)).(sin (2⁰).sin (88⁰)).........(sin (44⁰).sin (46⁰)).sin (45⁰)

So AB=(sin (1⁰).cos (1⁰)).(sin (2⁰).cos (2⁰)).........(sin (44⁰).cos (44⁰)).sin (45⁰)

So $2^{44}AB=\sin (2^0).\sin(4^0)........\sin(88^0).\frac{1}{\sqrt{2}}$

So $2^{44}AB = \frac{1}{\sqrt{2}}B$

So $A=\frac{1}{\sqrt{2}.2^{44}}=\frac{\sqrt{2}}{2^{45}}$

bcz B $\neq$ 0

[724] sakkmath

2012-05-18 15:35:51

The solution:

$\frac{\sqrt2}{2^{45}}$

For details, let me quote a little later.

Előzmény: [723] juantheron, 2012-05-18 07:06:44

[723] juantheron

2012-05-18 07:06:44

Thanks friends got it.

New task::

Numerical value of cos (1⁰).cos (3⁰).cos (5⁰).......cos (89⁰)=

where all angle are in Degree

[722] m2mm

2012-05-17 21:16:33

We may assume that x $\le$ y. Then (y+2)²=y²+4y+4>y²+3x>y². So, if y²+3x is a square, it must be (y+1)². Then y²+2y+1=y²+3x, so 2y+1=3x, which means that 2y $\equiv$ -1 (mod 3), so y $\equiv$ 1 (mod 3). Hence, y=3k+1 for some positive integer k. Since 2y+1=3x, then 6k+3=3x, 2k+1=x.

As x²+3y is a square, then (2k+1)²+3(3k+1)=4k²+4k+1+9k+3=4k²+13k+4 as well.

But (2k+2)²=4k²+8k+4 $\le$ 4k²+13k+4<4k²+16k+16=(2k+4)², so 4k²+13k+4=(2k+2)²=4k²+8k+4, or 4k²+13k+4=(2k+3)².

If 4k²+13k+4=4k²+8k+4, then k=0, x=y=1 (and this pair is a soultion).

If 4k²+13k+4=(2k+3)², then 4k²+13k+4=4k²+12k+9, so k=5. Then the solution is (11,16) or (16,11).

Előzmény: [717] juantheron, 2012-05-16 20:17:02

[721] logarlécész	2012-05-17 20:25:55
Teljesen igaz. Hiába, a figyelmetlenség mindig játszik. :-)
Előzmény: [712] nadorp, 2012-05-16 14:16:10

[720] nadorp

2012-05-17 14:18:36

"(x-y).(x+y-3)=(l+m).(l-m) means either x+y-3=l+m and x-y=l-m "

Incorrect. For example

3^.8=6^.4 but 8 $\neq$ 6 and 3 $\neq$ 4

Előzmény: [719] juantheron, 2012-05-17 08:49:02

[719] juantheron

2012-05-17 08:49:02

Thanks Nadorp got it

task:: find positive Integer value of x and y for which x²+3y and y²+3x is an perfect square

My solution::

Let x²+3y=l² and y²+3x=m²

Where l,m $\in$ Z

Then (x²-y²)-3(x-y)=l²-m²

(x-y).(x+y-3)=(l+m).(l-m)

means either x+y-3=l+m and x-y=l-m

2x=3+2l or $x=\frac{3+2l}{2}\notin Z$

similarly $y=\frac{3+2m}{2}\notin Z$

but when i put rendomly put values like (1,1) and (11,16) and (16,11)

but how can i calculate it

Thanks

[718] nadorp

2012-05-17 08:23:48

"But Not sure about k7 has no value for which satisfy the given equation"

Answer:

x²=4+[x] $\leq$ 4+x

x²-x-4 $\leq$ 0

$-\sqrt3<\frac{1-\sqrt{17}}2\leq x\leq\frac{1+\sqrt{17}}2<\sqrt7$

Előzmény: [716] juantheron, 2012-05-16 19:46:00

[717] juantheron

2012-05-16 20:17:02

Nwe task::

The number of ordered pairs of positive integers x,y such that x²+3y and y²+3x are both perfect squares

[716] juantheron

2012-05-16 19:46:00

Yes Friends your answer is Right

I have solved like that way

x²=4+[x]

here R.H.S(right hand side) is an Integer that means L.H.S(Left hand side must be an Integer)

So x² $\in$ Z

So Let x²=k, where k>0

So $x=\pm \sqrt{k}$

So our equation convert into $k=4+[\pm \sqrt{k}]$

Now after checking certain value of k like k=1,2,3,4,5,6

We Get only k=2 and k=6 satisfy the given equation

and after k $\geq$ 7, we get L.H.S # R.H.S

But Not sure about k $\geq$ 7 has no value for which satisfy the given equation

can anyone clear last line explanation

Thanks

[715] juantheron

2012-05-16 19:35:38

Thanks Nadorp and Small Potato

for Integration I have solve it after yours work

I have solved it like in that way

$\int \frac{1}{\sin^3 x+\cos^3 x}dx$

$\int\frac{\sin^2 x+\cos^2 x}{(\sin^3 x+\cos^3x)}dx$

$=\frac{1}{3}(\int\frac{\sin x+\cos x}{\sin^2 x-\sin x.\cos x+\cos^2 x}+\int\frac{2}{\sin x+\cos x}dx$

$-\frac{2}{3}\int \frac{(\sin x+\cos x)}{1+(\cos x-\sin x)^2}dx+\frac{2}{3}\int\frac{1}{\sin x+\cos x}dx$

Put (cos x-sin x)=tand (sin x+cos x)dx=-dt

and $\sin x+\cos x=\sqrt{2}.\sin (x+\frac{\pi}{4})$

after that we can calculate easily

[714] SmallPotato	2012-05-16 14:21:38
Lekéstem :-)
Előzmény: [712] nadorp, 2012-05-16 14:16:10

[713] SmallPotato

2012-05-16 14:21:13

Az [x] nem az x egész részét jelenti?

Ha azt, akkor a gyökök (grafikus alapú megoldással) $x_1=-\sqrt{2}$ (ekkor 2=4+(-2)) és $x_2=\sqrt{6}$ (ekkor 6=4+2).

Előzmény: [708] logarlécész, 2012-05-16 11:05:44

[712] nadorp	2012-05-16 14:16:10
Szerintem tört részt néztél és nem egész részt. Két megoldás van $\sqrt6$ és $-\sqrt2$ . A megoldáshoz csak annyi kell, hogy [x] $\leq$ x és x² egész.
Előzmény: [708] logarlécész, 2012-05-16 11:05:44