MHB Inequality with area of triangle

[Saitama]

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.

Any help is appreciated. Thanks!

 

[Opalg]

http://en.wikipedia.org/wiki/Weitzenb%F6ck%27s_inequality

 

[Saitama]

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
$$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=4s^2-2(ab+bc+ca)$$

 

[Opalg]

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.

 

[Saitama]

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?

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.

 

[Opalg]

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 $(a+b+c)^2 \leqslant 3(a^2+b^2+c^2).$

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.

 

[Saitama]

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?

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?

 

[Opalg]

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 $\sqrt s.$

 

[Saitama]

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?

 

[Opalg]

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?

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.

 

[Saitama]

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?