General Help for Inequalities (1 Viewer)

karnbmx · Jul 10, 2012

Hey guys,

Just wondering. What is a good way to tackle an inequality question ie. what are some fundamental rules that you should remember/make use of when doing questions. The reason I ask is:
1. My teacher is an idiot. She doesn't really teach us anything.
2. I am only aware of using and manipulating $(a-b)^2 \geq 0$ to answer questions. It gets me around some of them, but not most of them.

Thanks.

Shadowdude · Jul 10, 2012

Post an example question you're having trouble with?

Carrotsticks · Jul 10, 2012

Also good to know the AM-GM Inequality.

asianese · Jul 10, 2012

Along with $a^2+b^2\geq 2ab$ you can also use $(a+b)^2\geq4ab$ . The proof is

$(a-b)^2\geq 0 \\ \therefore a^2+b^2-2ab\geq0 \\ \therefore a^2+b^2\geq2ab \; $add 2ab to both sides$ \; \\ a^2+b^2+2ab\geq 4ab \\ \therefore (a+b)^2\geq 4ab$

karnbmx · Jul 10, 2012

Carrotsticks said:
Also good to know the AM-GM Inequality.

This one you mean? (a+b/2) >= sqrt(ab)

Carrotsticks · Jul 10, 2012

Yep but can be generalised.

seanieg89 · Jul 10, 2012

karnbmx said:
Hey guys,

Just wondering. What is a good way to tackle an inequality question ie. what are some fundamental rules that you should remember/make use of when doing questions. The reason I ask is:
1. My teacher is an idiot. She doesn't really teach us anything.
2. I am only aware of using and manipulating $(a-b)^2 \geq 0$ to answer questions. It gets me around some of them, but not most of them.

Thanks.

Nearly every inequality in the MX2 course can be deduced as a consequence from $(a-b)^2 \geq 0$ . The only exceptions I can think of that I have seen are the inequalities that come from the use of calculus.

The Matrix · Jul 10, 2012

Question: If n is a positive integer prove that $\sqrt[n]{n!}\geq \sqrt{n}$ .
Solution: Since n!=1.2.3...(n-1)n and n!=n(n-1)...3.2.1, $(n!)^2=(1\times n)(2\times (n-1))...(n\times 1).$
We need to prove that each of the above factors is greater or equal to n. Indeed, for $1\leq k\leq n$ , $k(n-k+1)-n=(n-k)(k-1)\geq 0$
$\therefore k(n-k+1)\geq n$
$\therefore (n!)^2\geq n^n$
$\therefore \sqrt[n]{n!}\geq \sqrt{n}$

This is another inequality you could get, it doesn't use the simple Cauchy–Schwarz inequality...

The Matrix · Jul 10, 2012

Can anyone have a go at doing this question?
Prove that $a^3+b^3+c^3=3abc$ if and only if $a=b=c$ or $a+b+c=0$ .

zeebobDD · Jul 10, 2012

IF a=b=c then LHS = 3c^3

RHS = 3c*c*c = 3c^3

Shadowdude · Jul 10, 2012

zeebobDD said:
IF a=b=c then LHS = 3c^3

RHS = 3c*c*c = 3c^3

Remember to fully prove this "if and only if" statement, you have to prove that IF a^3 + b^3 + c^3 = 3abc, then a=b=c.

seanieg89 · Jul 10, 2012

The Matrix said:
Can anyone have a go at doing this question?
Prove that $a^3+b^3+c^3=3abc$ if and only if $a=b=c$ or $a+b+c=0$ .

$LHS-RHS=\frac{1}{2}(a+b+c)((a-b)^2+(a-c)^2+(b-c)^2).$

From this the result is immediate.

Sanjeet · Jul 11, 2012

asianese said:
Along with $a^2+b^2\geq 2ab$ you can also use $(a+b)^2\geq4ab$ . The proof is

$(a-b)^2\geq 0 \\ \therefore a^2-b^2+2ab\geq0 \\ \therefore a^2+b^2\geq2ab \; $add 2ab to both sides$ \; \\ a^2+b^2+2ab\geq 4ab \\ \therefore (a+b)^2\geq 4ab$

Check your second line

asianese · Jul 11, 2012

Lol awks. You know what I meant =)

General Help for Inequalities (1 Viewer)

karnbmx

Member

Shadowdude

Cult of Personality

Carrotsticks

Retired

asianese

Σ

karnbmx

Member

Carrotsticks

Retired

seanieg89

Well-Known Member

The Matrix

Member

The Matrix

Member

zeebobDD

Member

Shadowdude

Cult of Personality

seanieg89

Well-Known Member

Sanjeet

Member

asianese

Σ

Users Who Are Viewing This Thread (Users: 0, Guests: 1)