Inequality with area of triangle

In summary: Yes, that's a good rule to follow. In general, when we apply AM-GM, we are trying to find the maximum or minimum value of a certain expression. If we include a constant factor in our expression, it will not affect the location of the maximum or minimum, so we can ignore it to simplify the calculation.
  • #1
Saitama
4,243
93
Problem:
If A is the area and 2s the sum of three sides of a triangle, then:

A)$A\leq \frac{s^2}{3\sqrt{3}}$

B)$A=\frac{s^2}{2}$

C)$A>\frac{s^2}{\sqrt{3}}$

D)None

Attempt:
From heron's formula:
$$A=\sqrt{s(s-a)(s-b)(s-c)}$$
From AM-GM:
$$\frac{s+(s-a)+(s-b)+(s-c)}{4}\geq \left(s(s-a)(s-b)(s-c)\right)^{1/4}$$
$$\Rightarrow A \leq \left(\frac{4s-a-b-c}{4}\right)^2$$
$$\Rightarrow A\leq \left(\frac{2s}{4}\right)^2$$
$$\Rightarrow A\leq \frac{s^2}{4}$$
But the correct answer is A. :confused:

Any help is appreciated. Thanks!
 
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  • #2
http://en.wikipedia.org/wiki/Weitzenb%F6ck%27s_inequality
 
Last edited:
  • #3
Opalg said:
Weitzenböck's_inequality

Edit. For some reason, that link does not seem to work. The url is en.wikipedia.org/wiki/Weitzenböck's_inequality.

The inequality says:
$$a^2+b^2+c^2 \geq 4\sqrt{3}A$$
But how to write it in terms of $s$?

I tried
$$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=4s^2-2(ab+bc+ca)$$
 
  • #4
Pranav said:
The inequality says:
$$a^2+b^2+c^2 \geq 4\sqrt{3}A$$
But how to write it in terms of $s$?

I tried
$$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=4s^2-2(ab+bc+ca)$$
Sorry, I thought that the inequality that I cited would lead to the one that you want. But in fact it doesn't seem to help.
 
  • #5
Opalg said:
Sorry, I thought that the inequality that I cited would lead to the one that you want. But in fact it doesn't seem to help.

Ah. How to proceed now? :confused:

I have one more question, why is it wrong to solve the problem the way I did in my attempt. I can't see anything wrong with it.
 
  • #6
Opalg said:
Sorry, I thought that the inequality that I cited would lead to the one that you want. But in fact it doesn't seem to help.
On second thoughts, I think that it does help. Use the fact that $0\leqslant (b-c)^2 + (c-a)^2 + (a-b)^2 = 2\bigl((a^2+b^2+c^2) - (bc+ca+ab)\bigr)$ to deduce that $(a+b+c)^2 \leqslant 3(a^2+b^2+c^2).$

Pranav said:
I have one more question, why is it wrong to solve the problem the way I did in my attempt.
There is nothing wrong with your proof. It's just that it is not a sufficiently strong result to do what the question asks for.
 
  • #7
Opalg said:
On second thoughts, I think that it does help. Use the fact that $0\leqslant (b-c)^2 + (c-a)^2 + (a-b)^2 = 2\bigl((a^2+b^2+c^2) - (bc+ca+ab)\bigr)$ to deduce that $(a+b+c)^2 \leqslant 3(a^2+b^2+c^2).$
So I have two inequalities:
$$3(a^2+b^2+c^2)\geq 12\sqrt{3}A$$
$$3(a^2+b^2+c^2)\geq 4s^2$$
I am not sure but subtracting the above two inequalities gives the right answer. But is it ok to subtract the inequalities? :confused:

There is nothing wrong with your proof. It's just that it is not a sufficiently strong result to do what the question asks for.
How do I check if the result is "sufficiently strong" or not? :confused:
 
  • #8
My suggestion of Weitzenböck's_inequality does not seem to work after all. Here is another method, using Heron's formula. Apply the GM-AM inequality to get $\sqrt[3]{(s-a)(s-b)(s-c)} \leqslant \frac13(s-a + s-b + s-c) = \frac13s.$ Raise both sides to the power $3/2$ to get $\sqrt{(s-a)(s-b)(s-c)} \leqslant \dfrac1{3\sqrt3}s^{3/2}.$ Then all you have to do is to multiply both sides by $\sqrt s.$
 
  • #9
Opalg said:
My suggestion of Weitzenböck's_inequality does not seem to work after all. Here is another method, using Heron's formula. Apply the GM-AM inequality to get $\sqrt[3]{(s-a)(s-b)(s-c)} \leqslant \frac13(s-a + s-b + s-c) = \frac13s.$ Raise both sides to the power $3/2$ to get $\sqrt{(s-a)(s-b)(s-c)} \leqslant \dfrac1{3\sqrt3}s^{3/2}.$ Then all you have to do is to multiply both sides by $\sqrt s.$

Isn't that what I did in my attempt with $\sqrt{s}$ and then applying AM-GM inequality? Why the two methods give different answers? :confused:
 
  • #10
Pranav said:
Isn't that what I did in my attempt with $\sqrt{s}$ and then applying AM-GM inequality? Why the two methods give different answers? :confused:
Yes, it the same method. The only difference is that you applied AM-GM to the product of all four factors in Heron's formula. It turns out that you get a stronger inequality if you apply it to just three of the four factors, and then multiply by the fourth factor $\sqrt s$ at the end.
 
  • #11
Opalg said:
Yes, it the same method. The only difference is that you applied AM-GM to the product of all four factors in Heron's formula. It turns out that you get a stronger inequality if you apply it to just three of the four factors, and then multiply by the fourth factor $\sqrt s$ at the end.

I see it now, thanks Opalg! :)

$s$ is a constant as per the problem, can I make it as a rule that while applying AM-GM inequality, I should not include the constant factors?
 

FAQ: Inequality with area of triangle

How is the area of a triangle calculated?

The area of a triangle is calculated by multiplying the base length by the height and dividing the result by 2. The formula for this is A = (b*h)/2, where A is the area, b is the base length, and h is the height.

What does inequality mean in the context of the area of a triangle?

Inequality in the context of the area of a triangle refers to the relationship between the sides and angles of a triangle. It means that the lengths of the sides and the measures of the angles are not equal, creating an unequal distribution of area within the triangle.

How does changing the side lengths of a triangle affect its area?

Changing the side lengths of a triangle will affect its area. If the base length or height of the triangle is increased, the area will also increase. However, if the base length or height is decreased, the area will decrease as well.

Can two triangles with the same area have different side lengths?

Yes, two triangles with the same area can have different side lengths. This is because there are multiple combinations of side lengths and height that can result in the same area for a triangle.

How does the shape of a triangle affect its area?

The shape of a triangle can affect its area in different ways. For example, a right triangle with the same base and height as an equilateral triangle will have a smaller area. This is because the equilateral triangle has all equal sides, while the right triangle has one shorter side.

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