Given the lengths of two sides of a triangle, is it possible to obtain (sharp) bounds for the average length of the sides of the triangle? The title says it all.

Given the lengths of two sides of a triangle, is it possible to obtain (sharp) bounds for the average length of the sides of the triangle?

Let the lengths of two sides of the triangle be $a$ and $b$.  Then the length $c$ of the third side is bounded, by the Triangle Inequality, as follows:
$$|a - b| < c < a + b.$$
Another way to state it is
$$\max(a, b, c) < a + b + c - \max(a, b, c)$$
implying
$$2\cdot\max(a, b, c) < a + b + c.$$
Consequently, a lower bound for the average length of the sides of the triangle is given by
$$\frac{2}{3}\cdot\max(a, b, c) < \frac{a + b + c}{3}.$$
My questions are as follows:


    
*
    
*Is it possible to obtain a sharper lower bound for $(a + b + c)/3$?
    
*Is it possible to obtain a sharp upper bound for $(a + b + c)/3$?


Thanks!

According to the law of cosines given $a,b,\gamma$ to be respectively two sides ($a \ge b$) and the angle in between you get that
$$c^2 = a^2+b^2-2ab\cos \gamma$$
so the average of the three sides is
$$\frac{a+b+c}{3} = \frac{a+b+\sqrt{a^2+b^2-2ab\cos \gamma}}{3}$$
Now note that


*

*the minimum of the average is when $\cos \gamma = 1 \Rightarrow$
$$\frac{a+b+c}{3} = \frac{a+b+\sqrt{a^2-2ab+b^2}}{3} = \frac{a+b+\sqrt{(a-b)^2}}{3} = \frac{2a}{3}$$

*the maximum is when $\cos \gamma = -1 \Rightarrow$
$$\frac{a+b+c}{3} = \frac{a+b+\sqrt{a^2+b^2+2ab}}{3} = \frac{a+b+\sqrt{(a+b)^2}}{3} = \frac{2(a+b)}{3}$$
so
$$\frac{2a}{3} \le \frac{a+b+c}{3} \le \frac{2(a+b)}{3}$$

