HSC 2013 MX2 Marathon (archive) (10 Viewers)

seanieg89 · Jan 11, 2014

Re: HSC 2014 4U Marathon

seanieg89 said:
$For $a,b,c >1$, show that:\\ \\ $\frac{3}{1+abc}\leq \frac{1}{1+a^3}+\frac{1}{1+b^3}+\frac{1}{1+c^3}.$$

Here is my solution based on calculus:

$Consider: \\$f(x,y,z):=\frac{1}{1+x^3}+\frac{1}{1+y^3}+\frac{1}{1+z^3}-\frac{3}{1+xyz}$\\ for $x,y,z>1.$ \\ \\ So $\frac{\partial f}{\partial x}=3(g(yz)-g(x^2))$\\ \\ where \\ \\ $g(t):=\frac{t}{(1+xt)^2}.$\\ \\ Calculus shows that $g$ is decreasing for $t>1$, no matter what $x$ is.\\ \\ Consequently, if we keep $y,z$ fixed and vary $x$, we have that $f$ is decreasing for $x\leq \sqrt{yz}$ and increasing for $x\geq \sqrt{yz}$. Similar results hold for $y,z$ by symmetry.\\ \\ By symmetry, we can assume $x\geq y \geq z.$ Hence, noting that $\sqrt{xy}\geq y \geq z$ and applying the above monotonicity result in $z$ and then $x$ yields:\\ \\ $f(x,y,z)\geq f(x,y,y)\geq f(y,y,y) = 0.$\\ \\ This completes the proof.$

seanieg89 · Jan 12, 2014

Re: HSC 2014 4U Marathon

$Let $a,b,c>0$ with $abc=8$. Prove that:\\ \\ $\frac{a^2}{\sqrt{(1+a^3)(1+b^3)}}+\frac{b^2}{ \sqrt{(1+b^3)(1+c^3)} }+\frac{c^2}{\sqrt{(1+c^3)(1+a^3)}}\geq \frac{4}{3}.$$

Sy123 · Jan 16, 2014

Re: HSC 2014 4U Marathon

$\\ $Let$ \ \ \ a, \ b , \ p, \ q > 0 \ \ \ \ $and$ \ \ \ \frac{1}{p} + \frac{1}{q} = 1 \\ \\ $Prove that$ \ \ \frac{a^p}{p} + \frac{b^q}{q} \geq ab$

Sy123 · Jan 17, 2014

Re: HSC 2014 4U Marathon

$\\ $The roots of the polynomial$ \\ \\ P(x) = x^3 - \alpha x^2 + \beta x - \gamma \\ \\ $are$ \ \ \alpha , \beta , \gamma \\ \\ $Find$ \ \alpha, \beta, \gamma$

Sy123 · Jan 17, 2014

Re: HSC 2014 4U Marathon

$\\ $Prove for positive$ \ a,b,c \\ \\ \sqrt{\frac{2a}{a+b}} + \sqrt{\frac{2b}{b+c}} + \sqrt{\frac{2c}{c+a}} \leq 3$

dunjaaa · Jan 17, 2014

Re: HSC 2014 4U Marathon

Using sum and product of roots and through some manipulation you get; α=-1,β=-1,γ=1. Also α, β, γ can all equal 0 as well.

dunjaaa · Jan 17, 2014

Re: HSC 2014 4U Marathon

Just started conics today, did the area proof only

Screen shot 2014-01-17 at 2.59.21 PM.png

seanieg89 · Jan 18, 2014

Re: HSC 2014 4U Marathon

$Let $(x_k)$ and $(y_k)$ be two decreasing real-valued sequences. \\ \\ Prove that:\\ \\ $n\sum_{k=1}^n x_ky_k \geq \left(\sum_{k=1}^n x_k\right)\left(\sum_{k=1}^n y_k\right).$$

RealiseNothing · Jan 18, 2014

Re: HSC 2014 4U Marathon

seanieg89 said:
$Let $(x_k)$ and $(y_k)$ be two decreasing real-valued sequences. \\ \\ Prove that:\\ \\ $n\sum_{k=1}^n x_ky_k \geq \left(\sum_{k=1}^n x_k\right)\left(\sum_{k=1}^n y_k\right).$$

'n' repetitions of the re-arrangement inequality.

Though I guess this is cheating.

seanieg89 · Jan 18, 2014

Re: HSC 2014 4U Marathon

RealiseNothing said:
'n' repetitions of the re-arrangement inequality.

Though I guess this is cheating.

Yeah, proving this is a bit easier than proving the rearrangement inequality.

RealiseNothing · Jan 18, 2014

Re: HSC 2014 4U Marathon

seanieg89 said:
$Let $(x_k)$ and $(y_k)$ be two decreasing real-valued sequences. \\ \\ Prove that:\\ \\ $n\sum_{k=1}^n x_ky_k \geq \left(\sum_{k=1}^n x_k\right)\left(\sum_{k=1}^n y_k\right).$$

Consider:

$n\sum_{k=1}^{n} x_ky_k - (\sum_{k=1}^nx_k)(\sum_{k=1}^ny_k)$

You end up with something like this:

$(nx_ny_n - \sum_{k=1}^n x_ny_k) + (nx_{n-1}y_{n-1} - \sum_{k=1}^n x_{n-1}y_k) + ... + (nx_1y_1 - \sum_{k=1}^n x_1y_k)$

Now if you separate the $nx_ny_k$ into $x_ny_k + ... + x_ny_k$ you can factor out the $x_n$ to get something like this:

$x_n(y_n-y_{n-1}) + x_n(y_n-y_{n-2}) + ... + x_1(y_1 - y_2)$

Now we apply a rule to the term in the brackets: $(y_m-y_p)$

1) If $m>p$ then we leave the term as it is.

2) If $m<p$ then we make the term $-(y_p-y_m)$

Now we will have two terms in the form:

$x_a(y_a-y_b)$

and

$-x_b(y_a-y_b)$

Where in all cases $a>b$ from the rule above.

There will always only be two terms of this form, no more, no less. Combining them thus gives:

$(x_a-x_b)(y_a-y_b)$

So all terms are $\geq 0$ as $a>b$ and thus:

$n\sum_{k=1}^{n} x_ky_k - (\sum_{k=1}^nx_k)(\sum_{k=1}^ny_k) \geq 0$

$n\sum_{k=1}^{n} x_ky_k \geq (\sum_{k=1}^nx_k)(\sum_{k=1}^ny_k)$

seanieg89 · Jan 18, 2014

Re: HSC 2014 4U Marathon

RealiseNothing said:
Consider:

$n\sum_{k=1}^{n} x_ky_k - (\sum_{k=1}^nx_k)(\sum_{k=1}^ny_k)$

You end up with something like this:

$(nx_ny_n - \sum_{k=1}^n x_ny_k) + (nx_{n-1}y_{n-1} - \sum_{k=1}^n x_{n-1}y_k) + ... + (nx_1y_1 - \sum_{k=1}^n x_1y_k)$

Now if you separate the $nx_ny_k$ into $x_ny_k + ... + x_ny_k$ you can factor out the $x_n$ to get something like this:

$x_n(y_n-y_{n-1}) + x_n(y_n-y_{n-2}) + ... + x_1(y_1 - y_2)$

Now we apply a rule to the term in the brackets: $(y_m-y_p)$

1) If $m>p$ then we leave the term as it is.

2) If $m<p$ then we make the term $-(y_p-y_m)$

Now we will have two terms in the form:

$x_a(y_a-y_b)$

and

$-x_b(y_a-y_b)$

Where in all cases $a>b$ from the rule above.

There will always only be two terms of this form, no more, no less. Combining them thus gives:

$(x_a-x_b)(y_a-y_b)$

So all terms are $\geq 0$ as $a>b$ and thus:

$n\sum_{k=1}^{n} x_ky_k - (\sum_{k=1}^nx_k)(\sum_{k=1}^ny_k) \geq 0$

$n\sum_{k=1}^{n} x_ky_k \geq (\sum_{k=1}^nx_k)(\sum_{k=1}^ny_k)$

I don't think cases are necessary but that looks about right. My solution:

$Since both sequences are decreasing:\\ \\ $(x_i-x_j)(y_i-y_j)\geq 0.$\\ \\ Summing, we get:\\ \\ $\sum_{i=1}^n\sum_{j=1}^n (x_i-x_j)(y_i-y_j)=2n\sum_{i=1}^n x_i y_i-2\left(\sum_{i=1}^n x_i\right)\left(\sum_{j=1}^n y_j\right)\geq 0$\\ \\ from which the claim follows immediately.$

RealiseNothing · Jan 18, 2014

Re: HSC 2014 4U Marathon

seanieg89 said:
I don't think cases are necessary but that looks about right. My solution:

$Since both sequences are decreasing:\\ \\ $(x_i-x_j)(y_i-y_j)\geq 0.$\\ \\ Summing, we get:\\ \\ $\sum_{i=1}^n\sum_{j=1}^n (x_i-x_j)(y_i-y_j)=2n\sum_{i=1}^n x_i y_i-2\left(\sum_{i=1}^n x_i\right)\left(\sum_{j=1}^n y_j\right)\geq 0$\\ \\ from which the claim follows immediately.$

Yer I think that's the same as my way just the reverse. You started from $(x_i-x_j)(y_i-y_j) \geq 0$ where as I worked towards that as the end result.

RealiseNothing · Jan 18, 2014

Re: HSC 2014 4U Marathon

seanieg89 said:
I don't think cases are necessary but that looks about right.

And by all cases I meant all terms. i.e. all terms had $a>b$ and thus were positive.

obliviousninja · Jan 19, 2014

Re: HSC 2014 4U Marathon

Any 2014ers actually do any of this?

dunjaaa · Jan 19, 2014

Re: HSC 2014 4U Marathon

I have no idea how to do any of these inequalities

anomalousdecay · Jan 19, 2014

Re: HSC 2014 4U Marathon

dunjaaa said:
I have no idea how to do any of these inequalities

Ok, think about this.

If you have two real numbers, and you do this:

$(a-b)^{2}$

What domain will the result be in?

RealiseNothing · Jan 19, 2014

Re: HSC 2014 4U Marathon

anomalousdecay said:
Ok, think about this.

If you have two real numbers, and you do this:

$(a-b)^{2}$

What domain will the result be in?

That compared to the inequalities we post is like comparing general maths to first year.

asianese · Jan 19, 2014

Re: HSC 2014 4U Marathon

Channeling Trebla here: Remember that 4U kids have only done 1 term of 4U. Probably complex numbers / integ or some of the easier topics. Please cater for them!! They'll get deterred otherwise (as above)!

Question:

$$If $ \omega $ is a non-real, primitive $n$-th root of unity, show that $ \\ (1-\omega)(1-\omega^2)\cdots(1-\omega^{n-1}) = n.$

obliviousninja · Jan 19, 2014

RealiseNothing said:
That compared to the inequalities we post is like comparing general maths to first year.

gg

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