question based on AM/GM

Question

If $a, b, c$ are positive real numbers and \begin{align*} a^{2}(1 + b^{2})+ b^{2}(1 + c^{2}) + c^{2}(1 + a^{2}) = 6abc \end{align*}

Find $a + b + c$.

Answer 1

On the one hand we have $1+a^2>= 2a$, $1+b^2 >= 2b$, and $1+c^2>= 2c$. On the other, from AM/GM we get $$a^2(1+b^2) + b^2(1+c^2) + c^2(1+a^2) >= 3((abc)^{2}(1+a^2)(1+b^2)(1+c^2))^{\frac{1}{3}}$$ Now, using three inequalities listed at the top: $$a^2(1+b^2) + b^2(1+c^2) + c^2(1+a^2) >= 3((abc)^{2}(1+a^2)(1+b^2)(1+c^2))^{\frac{1}{3}} >= 3((abc)^{2}2a2b2c)^{\frac{1}{3}} = 6abc$$ With equality only when we have equality in all inequalities above. So $1+a^2 = 2a$ which means $a=1$ and, similarly $b=1$ and $c=1$. Hence $a+b+c = 3$