Triangles Are Not Always Equilateral

The answer is $1.2$ . The equality case happens when we have a (degenerate) triangle with side lengths of $a = 2, b=1, c=1$ (and constant multiples). Note: If we want only non-degenerate triangles, then we do get strict inequality, though the largest $K$ value will still be 1.2. I didn't want to deal with such edge cases, hence set it up with an equality sign.

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First, notice that with $a = 2 , b = 1, c = 1$ , we get that $K \leq \frac{6}{5}$ . We will show that the largest possible value is $\frac{6}{5}$ . The inequalities that will be used are $b+c \geq a$ (triangle inequality) and $(b-c)^2 \geq 0$ (trivial inequality). It is clear that the equality case listed above satisfies both of these conditions.

First, since $b+c \geq a$ hence $a + b + c \geq 2a$ and so $bc \geq \frac{ 2abc } { a+b+c }$ .
Second, we have $a^2 \leq (b+c)^2 = b^2 + 2bc + c^2 \leq 2 b^2 + 2c^2$ .

Thus, it suffices to show that

$a^2 + b^2 + c^2 \geq \frac{ 6}{5} (a^2 + bc) \quad \left( \text{ which is } \geq \frac{6}{5} (a^2 + \frac{ 2abc}{a+b+c} ) \right) .$

This is equivalent to

$5 b^2 + 5c^2 \geq a^2 + 6 bc \Leftrightarrow 3 (b-c)^2 + 2 (b^2 + c^2) \geq a^2,$

which is true because $3 (b-c)^2 \geq 0$ and $2(b^2 + c^2 ) \geq a^2$ .

Since it is symmetric with respect to b and c, and b and c are restrained, it makes sense that $b=c$ . Let $b=c=1$ . Now plugging this into the given statement, we have that $\frac{(a^3+2a^2+2a+4)}{(a^3+2a^2+2a)} \ge K$ . This function is clearly decreasing, thus the maximum value of K is attained at the greatest value of a. Since $a<2b$ from the triangle inequality, we can plug in $a=2$ to get $\boxed{K=1.20}$