Triangle and logarithm problem

[Saitama]

Problem:
Given that $a,b$ and $c$ are the sides of $\Delta ABC$ such that $$z=\log_{2^a+2^{-a}} \left(4(ab+bc+ca)-(a+b+c)^2\right)$$ then $z$ has a real value if and only if

A)a=b=2c
B)3a=2b=c
C)a-b=3c
D)None of these

Attempt:
I am not sure where to start with this kind of problem. I wrote the expression inside the logarithm as follows:
$$-(a^2+b^2+c^2-2(ab+bc+ca))$$
Since, anything inside the logarithm must be greater than zero, I have
$$a^2+b^2+c^2-2(ab+bc+ca)<0 \Rightarrow a^2+b^2+c^2<2(ab+bc+ca)$$
But $a^2+b^2+c^2 >ab+bc+ca$, I get $ab+bc+ca>0$. I doubt this is going to help.

Any help is appreciated. Thanks!

 

[I like Serena]

Problem:
Given that $a,b$ and $c$ are the sides of $\Delta ABC$ such that $$z=\log_{2^a+2^{-a}} \left(4(ab+bc+ca)-(a+b+c)^2\right)$$ then $z$ has a real value if and only if

A)a=b=2c
B)3a=2b=c
C)a-b=3c
D)None of these

Attempt:
I am not sure where to start with this kind of problem. I wrote the expression inside the logarithm as follows:
$$-(a^2+b^2+c^2-2(ab+bc+ca))$$
Since, anything inside the logarithm must be greater than zero,

Hey Pranav! :)

I agree that the argument to the logarithm should be greater than zero.
From that point on the problem is symmetric in $a,b,c$.
That is, they can be swapped around giving the same expression.
In other words, none of the answers A, B, or C fit.
Therefore the answer must be D.

I have
$$a^2+b^2+c^2-2(ab+bc+ca)<0 \Rightarrow a^2+b^2+c^2<2(ab+bc+ca)$$
But $a^2+b^2+c^2 >ab+bc+ca$, I get $ab+bc+ca>0$. I doubt this is going to help.

Any help is appreciated. Thanks!

I believe that should be a $\ge$ sign.

Anyway, since $a,b,c$ are the sides of a triangle (which you did not use yet), you can conclude that this is always the case.

 

[Opalg]

If $a,b,c$ are the sides of a triangle then $b+c > a$, $c+a>b$ and $a+b>c$. Multiply the first of those inequalities by $a$, the second by $b$ and the third by $c$, and add. That gives $2(bc+ca+ab) > a^2+b^2+c^2.$ Therefore $4(bc+ca+ab) > a^2+b^2+c^2 + 2(bc+ca+ab) = (a+b+c)^2.$ So $4(bc+ca+ab) - (a+b+c)^2$ is always positive and therefore always has a real logarithm (to any base).

 

[Saitama]

That is, they can be swapped around giving the same expression.

Hi ILS! :)

I understand the above statement but how do you conclude the following:

In other words, none of the answers A, B, or C fit.

If $a,b,c$ are the sides of a triangle then $b+c > a$, $c+a>b$ and $a+b>c$. Multiply the first of those inequalities by $a$, the second by $b$ and the third by $c$, and add. That gives $2(bc+ca+ab) > a^2+b^2+c^2.$ Therefore $4(bc+ca+ab) > a^2+b^2+c^2 + 2(bc+ca+ab) = (a+b+c)^2.$ So $4(bc+ca+ab) - (a+b+c)^2$ is always positive and therefore always has a real logarithm (to any base).

Thanks a lot Opalg, very nicely done. :)

 

[I like Serena]

Hi ILS! :)

I understand the above statement but how do you conclude the following:

In other words, none of the answers A, B, or C fit.

Since the problem is symmetric in $a, b, c$ the correct answer will also have to be symmetric in $a, b, c$. Neither of the answers A, B, or C are.

 

[Saitama]

Since the problem is symmetric in $a, b, c$ the correct answer will also have to be symmetric in $a, b, c$. Neither of the answers A, B, or C are.

I see it now, thanks ILS! :)