Let $a$, $b$, $c$ be positive real numbers. Determine the minimal value of $$\frac{3a}{b+c}+\frac{4b}{c+a}+\frac{5c}{a+b}.$$
I have solved this problem using Chebyshev's and Nesbitt's inequalities.
The actual answer was $$\frac{1}{2}(\sqrt{3}+\sqrt{4}+\sqrt{5})^2 - 12$$ which is approximately 5.8. Help me understand where I went wrong.
In the image I have misspelt 'Nesbitt' as 'Nessbit'.