A geometry problem by Anish Roy

Let $f(a,b,c ) = \frac{ a^2 + 2bc } { b^2 + c^2 } + \frac{ b^2 + 2ca } { c^2 + a^2 } + \frac{ c^2 + 2ab } { a^2 + b^2 }$ .

From the triangle inequality, since $a > |b - c |$ , hence $a^2 > |b-c|^2 = b^2 - 2bc + c^2 \Rightarrow a^2 + 2bc > b^2 + c^2$ . Thus $\frac{ a^2 + 2bc } { b^2 + c^2 } > 1$ .
Since this is true of all 3 terms, this establishes that $f(a,b , c) > 3$ , so 3 is a lower bound.

Now, we see that we get equality when $a = |b-c|$ , which implies that we have a degenerate triangle. This leads us to believe that the infimum should be 3.

Observe that $f(a, b, c)$ is a continuous function in the variable $c > 0$ . Hence, since we know that $f(1, 1, 0) = 3$ , thus for any $\epsilon > 0$ , there exists a $0 < \delta < 2$ such that $| f( 1, 1, \delta ) - 3 | < \epsilon$ . This gives us $f(1, 1, \delta ) < 3 + \epsilon$ , indicating that any number larger than 3 is not a lower bound.

Hence, 3 is the greatest lower bound. It is an infimum (as opposed to a minimum) because equality cannot be achieved.