Applying AM-GM inequality into finding extreme/absolute value

1. AM-GM inequality

First, I would like to introduce standard AM-GM inequality:\[ x + y \geq 2\sqrt{xy} \quad \forall x,y \geq 0\]
Sign "=" occurs \(\iff x=y\)

Other version of AM-GM inequality is: \((x+y)^2 \geq 4xy \quad \forall x,y \in R\)

The extended version of AM-GM inequality:
\[x_1+x_2+...+x_n \geq n\sqrt[n]{x_1x_2...x_n} \quad \forall x_k \geq 0\]
Sign "=" occurs \(\iff x_1=x_2=...=x_n\)

In GMAT, for \(n=2\) and \(n=3\) are enough to solve every questions.

Note that sign \(\forall\) means for all
Sign \(\implies\) means imply. For example, \(a \implies b\) means \(a\) implies \(b\) but \(b\) doesnot imply \(a\)
Sign \(\iff\) means if and only if. For example, \(a \iff b\) means \(a\) occurs if and only if \(b\) occurs. This could also be expressed like this: \(a\) implies \(b\) and also \(b\) implies \(a\), or \(a \implies b\) and \(b \implies a\).

2. Proving AM-GM inequality

Since \(x,y \geq 0\), we have \(x + y \geq 2\sqrt{xy} \iff (x+y)^2 \geq 4xy \iff x^2 +2xy +y^2 \geq 4xy \iff (x-y)^2 \geq 0\)
The inequality \((x-y)^2 \geq 0\) is true for all \((x,y)\), so AM-GM inequality is true.

3. Applying AM-GM inequality into finding extreme value
Here are some examples to illustrate method of applying AM-GM inequality.


Example 1. For x and y are positive and x+y=10. Find the maximum value of xy.
Solve:
Since \(x,y>0\), \(x,y\) are satisfied to apply AM_GM inequality.

\(x+y \geq 2\sqrt{xy} \iff (x+y)^2 \geq 4xy \iff xy \leq \frac{(x+y)^2}{4} =\frac{10^2}{4}=25\)
\(max(xy)=25 \iff x=y=5\).


Example 2. For x and y are positive and xy=36. Find the minimum value of x+y.
https://gmatclub.com/forum/which-one-of- ... l#p1768552
Solve:
Since \(x,y>0\), \(x,y\) are satisfied to apply AM_GM inequality.
\(x+y \geq 2\sqrt{xy} = 2\sqrt{36} =12\)
\(min(x+y)=12 \iff x=y=6\).


Exmaple 3. For x and y are positive and 2x+3y=12. Find the maximum value of xy.
Solve:
Since \(x,y>0\), \(2x,3y\) are satisfied to apply AM_GM inequality.
\(2x+3y \geq 2\sqrt{2x \times 3y} = 2\sqrt{6xy} \iff xy \leq \frac{(2x+3y)^2}{24}=\frac{12^2}{24}=6\)
\(max(xy)=12 \iff 2x=3y=6 \iff x=3 \quad y=2\).


Example 4. Triangle ABC has perimeter 30. What is the lagest area of triangle ABC?
https://gmatclub.com/forum/patricia-buil ... l#p1766993

Solve:
Triangle ABC has 3 side lengths are \(a,b,c\).

Heron's formula to calculate triangle's area based on the lengths of 3 sides:
\(S_{ABC}=\sqrt{p(p-a)(p-b)(p-c)} \quad\) with \(p=\frac{a+b+c}{2}\)

We have \(p=30/2=15\).

Since \(p-a,p-b,p-c>0\), so \(p-a,p-b,p-c\) are satisfied to apply AM_GM inequality for 3 positive numbers.

The AM-GM inequality for 3 positive numbers: \(x+y+z \geq 3\sqrt[3]{xyz} \iff xyz \leq \frac{(x+y+z)^3}{27}\)

\(S_{ABC}=\sqrt{p(p-a)(p-b)(p-c)} \leq \sqrt{p \times \frac{(p-a+p-b+p-c)^3}{27}}=\sqrt{p \times \frac{p^3}{27}}=\) \(\sqrt{\frac{p^4}{27}}=\frac{p^2}{3\sqrt{3}}=25\sqrt{3}\)

\(max(S_{ABC})=25\sqrt{3} \iff a=b=c=10\) or ABC is equilateral triangle


4. Applying AM-GM inequality into solving equality with absolute
(in)equations with absolute are generally quite complicated to solve since the absolute value is based on specific value of expression. To solve these (in)equation, the simpliest way is to get rid of absolute site by specifying each cases. However, there are some kinds of (in)equation that could be solve quickly by using AM-GM inequality.


Example 5. Proving that \(|a|+|b| \geq |a+b| \quad \forall a,b \in R\).
Solve:
\(|a|+|b| \geq |a+b| \iff (|a|+|b|)^2 \geq (a+b)^2 \iff a^2 + 2|a||b| + b^2 \geq a^2 + 2ab + b^2 \iff |a||b| \geq ab\).
Since \(|a||b| \geq ab\) is true for all \((a,b)\), so the inquality is proved.

\(|a|+|b| = |a+b| \iff ab \geq 0\)


Example 6. Find the minium value of \(|x+3| + |x-7|\).
Solve:
\(|x+3| + |x-7| = |x+3| + |7-x| \geq |x+3+7-x|=10\)
\(min(|x+3| + |x-7|)=10 \implies (x+3)(7-x) \geq 0 \iff (x+3)(x-7) \leq 0 \implies -3 \leq x \leq 7\).

To solve inequilitys like \((x-a)(a-b)\leq 0\), \((x-a)(a-b)\geq 0\), I'll discuss in new post later.


Example 7. (Bunuel's questions) https://gmatclub.com/forum/inequality-an ... ml#p653853
Is \(|x+y|>|x-y|\)?
(1) |x| > |y|
(2) |x-y| < |x|

Solve:
\(|x+y|>|x-y| \iff |x+y|^2>|x-y|^2 \iff x^2+2xy+y^2>x^2-2xy+y^2 \iff 4xy>0 \iff xy>0\)
So solving this example is simply consider \(xy\) is positive or not.


Example 8. (Bunuel's questions) https://gmatclub.com/forum/inequality-an ... ml#p653870
Is \(r=s\)?
(1) \(-s<=r<=s\)
(2) \(|r|>=s\)

\((1) \implies s>0 \quad |r| \leq s\)
\((2) \implies |r| \geq s\)

Combine (1) and (2) \(\implies s>0 \quad and \, s \leq |r| \leq |s| \implies |r|=s \implies r=s \: or \: r=-s\). Answer is E.


Example 9. (Bunuel's questions) https://gmatclub.com/forum/inequality-an ... ml#p653783
\(|x+2|=|y+2|\) what is the value of \(x+y\)?
(1) \(xy<0\)
(2) \(x>2 \:\;\; y<2\)

Solve:
\(|x+2|=|y+2| \iff (x+2)^2=(y+2)^2 \iff x^2+4x+4=y^2+4y+4\) \(\iff x^2 +4x - y^2-4y=0 \iff (x-y)(x+y-4)=0\)

(1) If \(x=y \implies xy=x^2 \geq 0\). Hence we must have \(x+y=4\). Sufficient
(2) Clearly \(x \neq y \implies x+y=4\). Sufficient.
The answer is D.


The AM-GM inequality is so powerful to solve some kinds of inequalities as well as some kinds of equations. This tool could assist us in finding the solution immediately and saving a lot of time. I know that this tool is quite relevant to advanced mathematics and is somewhat hard to apply into actual GMAT questions. However, I believe that this post is useful for you in finding new (and maybe quick) ways to solve quant questions in GMAT.

Last but not least, the AM-GM inequality and its other versions are nothing but just tools to help us solve math quicker. The most important is that we need to specify what kind of each question is and dig for its solutions.