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rdj5933mile5math64
 Post subject: Ineq Questions that End Up Proving a Very Nice Ineq
PostPosted: Fri, 6 Apr 2012 00:37:42 UTC 
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Information needed to know:
Olympiad Level Inequalities such as:

AM-GM, Cauchy-Schwartz, Schur's, Rearrangement, Jensen's, Muirhead's

Problems:
1.Show that for all positive a,b,c:

\prod \frac{2a^2b}{c+a} \leq a^2b^2c^2

Spoiler:
It's equivalent to \frac{(a+b)(b+c)(c+a)}{8} \ge abc which is just AM-GM.



2.Prove that for all positive a,b,c:

\sum \frac{4a^2b^3c}{(c+a)(a+b)}

I wasn't sure how to do how to do this one.~


3.Prove that for all positive a,b,c:

\sum a^2 \ge \sum \frac{2a^2b}{a+c}

Darn couldn't get this one either. :(


4.Let x_1,x_2,x_3,y_1,y_2,y_3 be nonnegative reals such that

\sum x_i \ge \sum y_i

\sum x_ix_j \ge \sum y_iy_j

x_1x_2x_3 > y_1y_2y_3

Then prove: \sum \sqrt{x_i} \ge \sqrt{y_i}

Solution Sketch:

Spoiler:
Assume for the sake of contradiction that this is not the case.

Let \frac{x_1x_2x_3}{y_1y_2y_3}=n^2, where n>1.

Similarly define \frac{\sum \sqrt{b_i}}{\sum \sqrt{a_i}}=m, where m>1.

Let k be the unique positive integer satisfying n^k >m

What we wish to show is that:
\sum \sqrt{x_i} \ge \sum \sqrt{y_i}

Since both the RHS and LHS are positive, this is equivalent to:

\sum x_i +2\sum \sqrt{x_ix_j} \ge \sum y_i +2\sum \sqrt{y_iy_j}

So we just need to show that:

\sum \sqrt{x_ix_j} \ge \sum \sqrt{y_iy_j}

It suffices to show:

\sum x_ix_j +2\sqrt{x_1x_2x_3} \sum \sqrt{x_i} \ge \sum y_iy_j +2\sqrt{y_1y_2y_3} \sum \sqrt{y_i}

Thus, we want to prove that:

n \sum \sqrt{x_i} \ge \sum \sqrt{y_i}

Since the RHS of the above equation is greater than 0, the result follows iff:

n^2 \sum x_i +2n^2 \sum \sqrt{x_ix_j} \ge \sum y_i +2 \sum \sqrt{y_iy_j}

As a result, it suffices to show that:

n^2 \sum \sqrt{x_ix_j} \ge \sum \sqrt{y_iy_j}

The above inequality is true iff:

n^4 \sum x_ix_j +2n^4\sqrt{x_1x_2x_3} \sum \sqrt{x_i} \ge \sum y_iy_j + 2\sqrt{y_1y_2y_3} \sum \sqrt{y_i}

Or:

n^5 \sum \sqrt{x_i} \ge \sum \sqrt{y_i}

But, by repeating the above procedure the exponent of the n "explodes" since k is a fixed yet arbitrary positive integer it must be passed eventually.

This is a contradiction and our lemma is true.



5. Let a,b,c be positive real numbers. Prove that

\frac{ab}{\sqrt{ab+bc}}+\frac{bc}{\sqrt{bc+ca}}+\frac{ca}{\sqrt{ca+ab}}\le \frac{a+b+c}{\sqrt{2}}


It's supposed to work so that we prove 1-4 and then 5. Unfortunately I can't show 2 and 3.


Unrelated Inequality:

Suppose that the real numbers a_1,a_2,... a_{100} satisfy:

a_1 \ge a_2 \ge ... \ge a_{100} \ge 0

a_1+a_2 \leq 100

a_3+a_4+ ... +a_{100} \leq 100

Determine the maximum of:

a_1^2+a_2^2+ ... +a_{100}^2

and when equality occurs

EDIT:Typo

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Shadow
 Post subject: Re: Ineq Questions that End Up Proving a Very Nice Ineq
PostPosted: Fri, 6 Apr 2012 01:18:55 UTC 
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rdj5933mile5math64 wrote:
Information needed to know:
Olympiad Level Inequalities such as:

AM-GM, Cauchy-Schwartz, Schur's, Rearrangement, Jensen's, Muirhead's

Problems:
1.Show that for all positive a,b,c:

\prod \frac{2a^2b}{c+a} \leq a^2b^2c^2

Spoiler:
It's equivalent to \frac{(a+b)(b+c)(c+a)}{8} \ge abc which is just AM-GM.



2.Prove that for all positive a,b,c:

\sum \frac{4a^2b^3c}{(c+a)(a+b)}

I wasn't sure how to do how to do this one.~


3.Prove that for all positive a,b,c:

\sum a^2 \ge \sum \frac{2a^2b}{a+c}

Darn couldn't get this one either. :(


4.Let x_1,x_2,x_3,y_1,y_2,y_3 be nonnegative reals such that

\sum x_i \ge \sum y_i

\sum x_ix_j \ge \sum y_iy_j

x_1x_2x_3 > y_1y_2y_3

Then prove: \sum \sqrt{x_i} \ge \sqrt{y_i}

Solution Sketch:

Spoiler:
Assume for the sake of contradiction that this is not the case.

Let \frac{x_1x_2x_3}{y_1y_2y_3}=n^2, where n>1.

Similarly define \frac{\sum \sqrt{b_i}}{\sum \sqrt{a_i}}=m, where m>1.

Let k be the unique positive integer satisfying n^k >m

What we wish to show is that:
\sum \sqrt{x_i} \ge \sum \sqrt{y_i}

Since both the RHS and LHS are positive, this is equivalent to:

\sum x_i +2\sum \sqrt{x_ix_j} \ge \sum y_i +2\sum \sqrt{y_iy_j}

So we just need to show that:

\sum \sqrt{x_ix_j} \ge \sum \sqrt{y_iy_j}

It suffices to show:

\sum x_ix_j +2\sqrt{x_1x_2x_3} \sum \sqrt{x_i} \ge \sum y_iy_j +2\sqrt{y_1y_2y_3} \sum \sqrt{y_i}

Thus, we want to prove that:

n \sum \sqrt{x_i} \ge \sum \sqrt{y_i}

Since the RHS of the above equation is greater than 0, the result follows iff:

n^2 \sum x_i +2n^2 \sum \sqrt{x_ix_j} \ge \sum y_i +2 \sum \sqrt{y_iy_j}

As a result, it suffices to show that:

n^2 \sum \sqrt{x_ix_j} \ge \sum \sqrt{y_iy_j}

The above inequality is true iff:

n^4 \sum x_ix_j +2n^4\sqrt{x_1x_2x_3} \sum \sqrt{x_i} \ge \sum y_iy_j + 2\sqrt{y_1y_2y_3} \sum \sqrt{y_i}

Or:

n^5 \sum \sqrt{x_i} \ge \sum \sqrt{y_i}

But, by repeating the above procedure the exponent of the n "explodes" since k is a fixed yet arbitrary positive integer it must be passed eventually.

This is a contradiction and our lemma is true.



5. Let a,b,c be positive real numbers. Prove that

\frac{ab}{\sqrt{ab+bc}}+\frac{bc}{\sqrt{bc+ca}}+\frac{ca}{\sqrt{ca+ab}}\le \frac{a+b+c}{\sqrt{2}}


It's supposed to work so that we prove 1-4 and then 5. Unfortunately I can't show 2 and 3.


Unrelated Inequality:

Suppose that the real numbers a_1,a_2,... a_{100} satisfy:

a_1 \ge a_2 \ge ... \ge a_{100} \ge 0

a_1+a_2 \leq 100

a_3+a_4+ ... +a_{100} \leq 100

Determine the maximum of:

a_1^2+a_2^2+ ... +a_{100}^2

and when equality occurs

EDIT:Typo


I'm not sure "Olympiad level" is something that can be said for inequalities--AM-GM and almost all the others you listed are extremely standard. Perhaps you mean the application is somehow nonstandard or more difficult to see.

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rdj5933mile5math64
 Post subject: Re: Ineq Questions that End Up Proving a Very Nice Ineq
PostPosted: Fri, 6 Apr 2012 01:24:18 UTC 
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Olympiad level for high schoolers.~ :P

EDIT: what I mean by olympiad level is high school mathematical proof contests similar in difficulty to the usamo

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outermeasure
 Post subject: Re: Ineq Questions that End Up Proving a Very Nice Ineq
PostPosted: Fri, 6 Apr 2012 04:34:20 UTC 
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rdj5933mile5math64 wrote:
2.Prove that for all positive a,b,c:

\sum \frac{4a^2b^3c}{(c+a)(a+b)}

I wasn't sure how to do how to do this one.~


I don't know either. What am I supposed to prove with \sum\frac{4a^2b^3c}{(c+a)(a+b)}?

rdj5933mile5math64 wrote:
3.Prove that for all positive a,b,c:

\sum a^2 \ge \sum \frac{2a^2b}{a+c}

Darn couldn't get this one either. :(


Assuming \sum means cyclic sum, not symmetric sum:
Spoiler:
Move everything to the LHS and clearing denominators, it suffices to prove
\sum_{\text{symmetric}}a^4b-\sum_{\text{symmetric}}a^3b^2\geq 0
which is just an instance of the Muirhead's inequality.

For more about Muirhead's inequality, consult wikipedia/AoPS/... It is an inequality that trivialises a very large proportion of olympiad inequalities.

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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rdj5933mile5math64
 Post subject: Re: Ineq Questions that End Up Proving a Very Nice Ineq
PostPosted: Fri, 6 Apr 2012 05:47:18 UTC 
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outermeasure wrote:
Assuming \sum means cyclic sum, not symmetric sum:


That's correct I should have been more clear.

outermeasure wrote:
rdj5933mile5math64 wrote:
2.Prove that for all positive a,b,c:

\sum \frac{4a^2b^3c}{(c+a)(a+b)}

I wasn't sure how to do how to do this one.~


I don't know either. What am I supposed to prove with \sum\frac{4a^2b^3c}{(c+a)(a+b)}?


ehehe :oops: sorry the correct inequality is below

\sum \frac{4a^2b^3c}{(c+a)(a+b)} \leq \sum a^2b^2


outermeasure wrote:
rdj5933mile5math64 wrote:
3.Prove that for all positive a,b,c:

\sum a^2 \ge \sum \frac{2a^2b}{a+c}

Darn couldn't get this one either. :(


Spoiler:
Move everything to the LHS and clearing denominators, it suffices to prove
\sum_{\text{symmetric}}a^4b-\sum_{\text{symmetric}}a^3b^2\geq 0
which is just an instance of the Muirhead's inequality.

For more about Muirhead's inequality, consult wikipedia/AoPS/... It is an inequality that trivialises a very large proportion of olympiad inequalities.


Awesome! Thank You! :D

And while I'm at it:
rdj5933mile5math64 wrote:
Unrelated Inequality:

Suppose that the real numbers a_1,a_2,... a_{100} satisfy:

a_1 \ge a_2 \ge ... \ge a_{100} \ge 0

a_1+a_2 \leq 100

a_3+a_4+ ... +a_{100} \leq 100

Determine the maximum of:

a_1^2+a_2^2+ ... +a_{100}^2

and when equality occurs


Solution Sketch:
Spoiler:
Multiplying the second given inequality by a_1 and using the first inequality yields:

a_1^2+a_2^2 \leq a_1+a_2 \leq 100a_1

Multiplying the third given inequality by a_3 and using the first inequality yields:

a_3^2+a_4^2+....+a_{100}^2 \leq 100a_3

So, we add the 2 inequalities to get that:

a_1^2+a_2^2+a_3^2+..... +a_{100}^2 \leq 100a_3+100a_1 \leq 100a_2+100a_1 \leq 10,000

Note that the maximum is reached when the inequalites turn into equalities so
a_1^2+a_2^2=a_1^2+a_2a_1=100a_1.

From this we get that
a_2(a_1-a_2)=0

a_1(100-a_1-a_2)=0

So, by the first equation either a_2=0 which means that a_1=100 or a_1=a_2=50.

Now, let i be a positive integer such that 4 \leq i \leq 100. Observe that: a_3(a_3-a_i)=0 and a_3+a_4+...+a_{100}=100

Also note that a_2=a_3. So, if a_2=0, then by the first given inequality everything else is 0. If a_2=50, then a_1=a_3=a_4=50

So equality holds when we have (100,0,0,0,....,0,0) or (50,50,50,50,0,0,....,0,0)

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