I have been working on this journal problem for my Undergraduate Summer Research for a couple weeks now and I'm stuck. The problem is:
Let $\triangle ABC$ be a triangle with its greatest angle less than $120^{\circ}$, and let $F$ be the Fermat point of $\triangle ABC$ (the point in the interior of $\triangle ABC$ that minimizes the sum of the distances to A, B, and C). We've to prove $$ \frac{FA^4}{AB^2} + \frac{FB^4}{BC^2} + \frac{FC^4}{CA^2} \geq \frac{FA^3+FB^3+FC^3}{FA+FB+FC}. $$
My progress so far:
After working on this problem for a while, I've found and/or proved some properties of the Fermat point of a triangle:
Definition: If $\triangle ABC$ has all angles less than $2\pi/3$, then the circumscribed circles of the equilateral triangles $\triangle ABC_1$, $\triangle ACB_1$, and $\triangle BCA_1$ exterior to $\triangle ABC$ have a common point. This point is called the Torricelli point of $\triangle ABC$.
Lemma-$1:$ For any triangle $\triangle ABC$, under the given conditions of the problem, the Fermat point and the Torricelli point coincide. Thus, $F$ is also called the Fermat-Torricelli point of $\triangle ABC$.
Lemma-$2:$ If $F$ is the Fermat-Torricelli point of $\triangle ABC$ then we have $$ FB+FC\leq \frac{2}{\sqrt{3}}BC,\\ FC+FA\leq \frac{2}{\sqrt{3}}CA,\\ FA+FB\leq \frac{2}{\sqrt{3}}AB .$$
Adding these three equations, we also obtain $$ FA+FB+FC\leq\frac{1}{\sqrt{3}}(AB+BC+CA) .$$
Lemma-$3:$ Under the given conditions in the problem for $\triangle ABC$, we have $$ FB^2+FC^2 \geq\frac{2}{3}BC^2,\\ FC^2+FA^2 \geq\frac{2}{3}CA^2,\\ FA^2+FB^2 \geq\frac{2}{3}AB^2. $$
Again, adding these equations we obtain $$ FA^2+FB^2+FC^2 \geq\frac{1}{3}(AB^2+BC^2+CA^2) .$$
Now, using Bergström's Inequality and Lemma-$3$, I manipulated the L.H.S. as follows $$ \frac{FA^4}{AB^2} + \frac{FB^4}{BC^2} + \frac{FC^4}{CA^2} \geq \frac{\left(FA^2+FB^2+FC^2 \right)^2}{AB^2+BC^2+CA^2}\geq \frac{1}{9}(AB^2+BC^2+CA^2) .$$
For the R.H.S., however, I could not get anywhere. My target was to conclude the proof by showing $$ R.H.S.\leq \frac{1}{9}(AB^2+BC^2+CA^2).$$ But that doesn't seem to be possible. I tried using Lemma-$2$ along with the factorized form of $x^3+y^3+z^3$ without success. Can someone provide some hints or any alternate approach for this problem? TIA.