1. ## Help

A triangle ABC,a=AC, b=AC, c=AB , and the
angle C≥60°

2. Why no one answer me?
Do I fail to express the question?

3. I want to help you but I do not understand your question, sorry. Can you draw it?

4. a triangle:......A..
.........................
................b.............c
.................................
...............C ...B
............................a

I hope you can understand my question now.
Looking forward to your help,thank you.

5. Is it isoseles? Because in the picture it is not and in the first post it is. If it is isosoles I have found a proof.

6. Yes

7. If I understand the problem correctly the angle can be any angle not just more than 60 and the inequality is really and equality. Let me explain:

Since $a=b$ the problem reduces to proving
$2a(\frac{2}{a}+\frac{1}{c}) \geq 4+\frac{1}{\sin (C/2)}$. Thus,
$4+\frac{2a}{c}\geq 4+\frac{1}{\sin (C/2)}$. Thus, (if and only if)
$\frac{2a}{c}\geq \frac{1}{\sin C/2}$.
But this is always equal to eachother, thus,
$\frac{2a}{c}= \frac{1}{\sin (C/2)}$.

Because if an isoseles triangle has sides $a$ with angle between them $C$ and $c$ is its third side then $2a\sin (C/2)=c$.
The reason why is simple, by the law of cosines, $a^2+a^2-2a^2\cos C=c^2$ thus, $2a^2(1-\cos C)=c^2$ by the half-angle identity $4a^2\sin^2(C/2)=c^2$ thus $2a\sin(c/2)=a$ thus, $\frac{2a}{c}=\frac{1}{\sin(C/2)}$. Thus your inequality is in reality always equal, thus the limitation for the angle were not needed.
Thus, your problem was tricky in that it was making me believe the angle needs to be greater than 60 which is not true.

Here are some examples which show what you are saying is an equality. A right triangle with sides $1,1,\sqrt{2}$, here the angle is 90. An equilateral triangle with sides $1,1,1$ here the angle is 60.

8. Thank you for your help.
But after this question, there is another one asking me to guess when a and b are not equal, can it still be right. I've put many sets of numbers in it, which shows it is always right. But I can't prove it.
I think I have to trouble again.

just like this:
A triangle ABC,a=BC, b=AC, c=AB , and the
angle C≥60°

9. I am trying to solve your problem for all triangles but I did not find a proof yet. Are you certain it works for cases that you tried it for? Where did you even get this problem?
Any ideas which I could use in my proof?

10. Now I'm sure that when angle C≥60°,it is always the case.
I've looked for this problem in many reference books, one of which says that it is true, but it doesn't offer a proof.

A triangle ABC,a=BC, b=AC, c=AB , and the
angle C≥60°