Author Topic: TT's Maths Thread (Read 133279 times) Tweet Share

TrueTears · « **Reply #180 on:** November 26, 2009, 02:14:54 am »

Quote from: kamil9876 on November 26, 2009, 01:19:29 am

consider the sequence:

1,2,3,4.....n

it's arithmetic mean is $\frac{\sum_{i=1}^{n}i}{n}=\frac{n+1}{2}$

It's geometric mean is: $(n!)^{\frac{1}{n}}$

GM<AM by AM-GM. From which the result follows.

I noticed this inequality has been used many times in this thread, I love this pic btw, saw it in a textbook(a better version tho )

(Image removed from quote.)

haha that was so trivial lol

Yeah that pic is awesome, using similar triangles to derive AM-GM.

Ahmad · « **Reply #181 on:** November 26, 2009, 02:43:10 am »

That picture is indeed cool!

Hint for 2: square everything then apply simple inequalities.

TrueTears · « **Reply #182 on:** November 26, 2009, 04:47:28 pm »

Quote from: kamil9876 on November 26, 2009, 01:56:23 am

q3.)

consider the sequence of $(x+a_1), (x+a_2).... (x+a_n)$ do some AM-GM.

lol thanks I got it.

Using AM-GM on that sequence yields:

$\frac{(x+a_1)+(x+a_2)+(x+a_3)+...+(x+a_n)}{n} \ge \left[(x+a_1)(x+a_2)(x+a_3)...(x+a_n)\right]^{\frac{1}{n}}$

$\frac{xn + a_1+a_2+a_3+...+a_n}{n} \ge \left[(x+a_1)(x+a_2)(x+a_3)...(x+a_n)\right]^{\frac{1}{n}}$

$\left(x + \frac{a_1+a_2+a_3+...+a_n}{n}\right) \ge \left[(x+a_1)(x+a_2)(x+a_3)...(x+a_n)\right]^{\frac{1}{n}}$

$\left(x + \frac{a_1+a_2+a_3+...+a_n}{n}\right)^n \ge (x+a_1)(x+a_2)(x+a_3)...(x+a_n)$ as required.

Quote from: Ahmad on November 26, 2009, 02:43:10 am

That picture is indeed cool!

Hint for 2: square everything then apply simple inequalities.

Thanks Ahmad! I think I got it now

TrueTears · « **Reply #183 on:** November 26, 2009, 10:05:45 pm »

Induction can be used for Q 2 but it is very tedious and not very elegant...

Quote

Show that $\frac{1}{\sqrt{4n}} \le \left(\frac{1}{2}\right)\left(\frac{3}{4}\right) ...\left(\frac{2n-1}{2n}\right) < \frac{1}{\sqrt{2n}}$

For lower bound:

Base Case

Let $n = 2$

$\left(\frac{1}{2}\right)\left(\frac{3}{4}\right) \ge \frac{1}{\sqrt{8}}$

Inductive Hypothesis

Assume it is true for $n = k$

$\frac{1}{\sqrt{4k}} \le \left(\frac{1}{2}\right)\left(\frac{3}{4}\right) ...\left(\frac{2k-1}{2k}\right)$ is true.

Inductive Step

Need to prove it is true for $n = k+1$

$\frac{1}{\sqrt{4(k+1)}} \le \left(\frac{1}{2}\right)\left(\frac{3}{4}\right) ...\left(\frac{2(k+1)-1}{2(k+1)}\right)$

Multiply Inductive Hypothesis by $\left(\frac{2(k+1)-1}{2(k+1)}\right)$

$\left(\frac{2(k+1)-1}{2(k+1)}\right)\left(\frac{1}{\sqrt{4k}}\right) \le \left(\frac{1}{2}\right)\left(\frac{3}{4}\right) ...\left(\frac{2k-1}{2k}\right)\left(\frac{2(k+1)-1}{2(k+1)}\right)$

If we can show $\left(\frac{2(k+1)-1}{2(k+1)}\right)\left(\frac{1}{\sqrt{4k}}\right) > \frac{1}{\sqrt{4(k+1)}}$ Then our proof is complete

$\mbox{LHS} = \left(\frac{2k+1}{2k+2}\right)\left(\frac{1}{\sqrt{4k}}\right) = \left(\frac{1}{\sqrt{4(k+1)}}\right)\left[\frac{2(2k+1)}{\sqrt{k+1}}\right]\left(\frac{1}{\sqrt{4k}}\right)$

Thus $\mbox{LHS} > \mbox{RHS}$ (Since $k > 1$ )

$\therefore \frac{1}{\sqrt{4n}} \le \left(\frac{1}{2}\right)\left(\frac{3}{4}\right) ...\left(\frac{2n-1}{2n}\right)$

For upper bound:

Base Case

Let $n = 2$

$\left(\frac{1}{2}\right)\left(\frac{3}{4}\right) < \frac{1}{2}$

Inductive Hypothesis:

Assume it is true for $n = k$

$\left(\frac{1}{2}\right)\left(\frac{3}{4}\right) ...\left(\frac{2k-1}{2k}\right) < \frac{1}{\sqrt{2k}}$ is true.

Inductive Step:

Need to prove it is true for $n = k+1$

$\left(\frac{1}{2}\right)\left(\frac{3}{4}\right) ...\left(\frac{2(k+1)-1}{2(k+1)}\right) < \frac{1}{\sqrt{2(k+1)}}$

Multiply Inductive Hypothesis by $\left(\frac{2(k+1)-1}{2(k+1)}\right)$

$\left(\frac{1}{2}\right)\left(\frac{3}{4}\right) ...\left(\frac{2k-1}{2k}\right)\left(\frac{2(k+1)-1}{2(k+1)}\right) < \left(\frac{1}{\sqrt{2k}}\right)\left(\frac{2(k+1)-1}{2(k+1)}\right)$

If we can show $\left(\frac{1}{\sqrt{2k}}\right)\left(\frac{2(k+1)-1}{2(k+1)}\right) < \frac{1}{\sqrt{2(k+1)}}$ then our proof is complete.

$\mbox{LHS} = \left(\frac{1}{\sqrt{2(k+1)}}\right)\left[\frac{\sqrt{2}(2k+1)}{2\sqrt{k+1}}\right]\left(\frac{1}{\sqrt{2k}}\right)$

Since $k > 1$ then $\mbox{LHS} < \mbox{RHS}$

$\therefore \left(\frac{1}{2}\right)\left(\frac{3}{4}\right) ...\left(\frac{2n-1}{2n}\right) < \frac{1}{\sqrt{2n}}$

Overall combining the lower and upper bounds we have proved: $\frac{1}{\sqrt{4n}} \le \left(\frac{1}{2}\right)\left(\frac{3}{4}\right) ...\left(\frac{2n-1}{2n}\right) < \frac{1}{\sqrt{2n}}$

TrueTears · « **Reply #184 on:** November 26, 2009, 10:25:22 pm »

Alright last set of questions and then combinatorics time!

~~1. Let $a,b,c,d \ge 0$ prove that $\frac{1}{a}+\frac{1}{b} + \frac{4}{c}+\frac{16}{d} \ge \frac{64}{a+b+c+d}$~~

~~2. Let $x,y,z \ge 0$ with $xyz = 1$ . Find the minimum value of $\frac{x}{y+z} + \frac{y}{x+z} + \frac{z}{x+y}$~~

~~3. Let $a,b,c \ge 0$ . Prove that $\sqrt{3(a+b+c)} \ge \sqrt{a} + \sqrt{b} + \sqrt{c}$~~

/0 · « **Reply #185 on:** November 26, 2009, 11:40:27 pm »

2. Looks a lot like http://en.wikipedia.org/wiki/Nesbitt%27s_inequality
But the constraint is different. Hmmm.

3.

Taking squares of both sides,

$a+b+c \geq \sqrt{ab}+\sqrt{bc}+\sqrt{ac}$

From am-gm

$\frac{a+b}{2} \geq \sqrt{ab}$

$\frac{b+c}{2} \geq \sqrt{bc}$

$\frac{a+c}{2} \geq \sqrt{ac}$

Adding:

$a+b+c \geq \sqrt{ab}+\sqrt{bc}+\sqrt{ac}$

Ahmad · « **Reply #186 on:** November 26, 2009, 11:46:08 pm »

It's also an application of CS,

$(1 + 1 + 1)(a + b + c) \ge (\sqrt{a} + \sqrt{b} + \sqrt{c})^2$

TrueTears · « **Reply #187 on:** November 26, 2009, 11:57:24 pm »

Quote from: /0 on November 26, 2009, 11:40:27 pm

2. Looks a lot like http://en.wikipedia.org/wiki/Nesbitt%27s_inequality
But the constraint is different. Hmmm.

3.

Taking squares of both sides,

$a+b+c \geq \sqrt{ab}+\sqrt{bc}+\sqrt{ac}$

From am-gm

$\frac{a+b}{2} \geq \sqrt{ab}$

$\frac{b+c}{2} \geq \sqrt{bc}$

$\frac{a+c}{2} \geq \sqrt{ac}$

Adding:

$a+b+c \geq \sqrt{ab}+\sqrt{bc}+\sqrt{ac}$

haha I actually just did that but I took the AM-GM of $a+b+c$ and compared it with $\sqrt{ab}+\sqrt{bc}+\sqrt{ac}$

/0 · « **Reply #188 on:** November 27, 2009, 12:03:41 am »

Quote from: TrueTears on November 26, 2009, 11:57:24 pm

Quote from: /0 on November 26, 2009, 11:40:27 pm
2. Looks a lot like http://en.wikipedia.org/wiki/Nesbitt%27s_inequality
But the constraint is different. Hmmm.

3.

Taking squares of both sides,

$a+b+c \geq \sqrt{ab}+\sqrt{bc}+\sqrt{ac}$

From am-gm

$\frac{a+b}{2} \geq \sqrt{ab}$

$\frac{b+c}{2} \geq \sqrt{bc}$

$\frac{a+c}{2} \geq \sqrt{ac}$

Adding:

$a+b+c \geq \sqrt{ab}+\sqrt{bc}+\sqrt{ac}$
haha I actually just did that but I took the AM-GM of $a+b+c$ and compared it with $\sqrt{ab}+\sqrt{bc}+\sqrt{ac}$

$a+b+c \geq 3\sqrt[3]{abc}$

$\sqrt{ab}+\sqrt{bc}+\sqrt{ac} \geq 3\sqrt[3]{\sqrt{ab}\sqrt{bc}\sqrt{ac}} = 3\sqrt[3]{abc}$ ?

Anyway got a bit further with 2

$\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}$

If we use $x = \frac{1}{yz}$ on the denominator terms and simplify we get

$\frac{x}{xz+xy} +...$

So we need to minimize

$\frac{1}{y+z} + \frac{1}{x+z} + \frac{1}{x+y}$

~~Then we could say $t = x+y+z$ so we need to minimize $\frac{1}{t-x}+\frac{1}{t-y}+\frac{1}{t-z}$ .....~~ nevermind shit idea

hmm

TrueTears · « **Reply #189 on:** November 27, 2009, 12:13:34 am »

I think I got Q 1

Some rearranging leads to:

$\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\right)\left(a+b+c+d\right) \ge (1+1+2+4)^2$

$\left[\left(\frac{1}{\sqrt{a}}\right)^2\left(\frac{1}{\sqrt{b}}\right)^2\left(\frac{2}{\sqrt{c}}\right)^2\left(\frac{4}{\sqrt{d}}\right)^2\right]\left(\sqrt{a}^2+\sqrt{b}^2+\sqrt{c}^2+\sqrt{d}^2\right) \ge (1+1+2+4)^2$

Which is true by Cauchy-Schwarz.

Ahmad · « **Reply #190 on:** November 27, 2009, 12:21:44 am »

The only part of the constraint in 2 that's important is that it means none of x, y, z can be 0. So it really does reduce to Nesbitt's, which is well known and admits many proofs

/0 · « **Reply #191 on:** November 27, 2009, 12:32:50 am »

Quote from: Ahmad on November 27, 2009, 12:21:44 am

The only part of the constraint in 2 that's important is that it means none of x, y, z can be 0. So it really does reduce to Nesbitt's, which is well known and admits many proofs

Oh right indeed! *facepalm*
For some reason I thought you needed $x+y+z=1$ for nesbitt

Ahmad · « **Reply #192 on:** November 27, 2009, 12:37:17 am »

Nope, but you can assume that anyway since the inequality is homogeneous, so you might've seen people exploiting that.

TrueTears · « **Reply #193 on:** November 27, 2009, 12:39:05 am »

Quote from: Ahmad on November 27, 2009, 12:37:17 am

Nope, but you can assume that anyway since the inequality is homogeneous, so you might've seen people exploiting that.

What does it mean when you say the inequality is homogeneous? As in it's 'symmetrical'? =]

Ahmad · « **Reply #194 on:** November 27, 2009, 01:27:57 am »

Suppose the inequality is $f(x,y,z) \ge 0$ then it's homogeneous if $f(x,y,z) = f(kx,ky,kz)$ .

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