Another Inequality

Question

Let $a$, $b$ and $c$ be positive reals such that $$\dfrac{a^3+2a}{a^2+1}+\dfrac{b^3+2b}{b^2+1}+\dfrac{c^3+2c}{c^2+1}=\dfrac 92,$$
then is it true that $\dfrac 1a+\dfrac1b+\dfrac1c\ge3$ ?

Answer 1

You can easily solve the question if you transform the problem in a nonlinear programming. This can be done by posing $$\min \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$ subject to $$a^3+2a)/(a^2+1)+(b^3+2b)/(b^2+1)+(c^3+2c)/(c^2+1)-9/2=0$$.

Then you can check if this minimum is greather than 3. This is a NLP in standard form, if you apply the first order necessary condition you find out that $a=b=c=1$ is a candidate minimum point, which yields a value of 3. Then you can apply the second order sufficient conditions and obtain that $(1,1,1)$ is really a minumum.

Answer 2

$$\sum_{cyc}\frac{1}{a}-3=\sum_{cyc}\left(\frac{1}{a}-1\right)=\sum_{cyc}\left(\frac{1}{a}-1+\frac{a^3+2a}{a^2+1}-\frac{3}{2}\right)=$$ $$=\sum_{cyc}\frac{(a-1)^2(2a^2-a+2)}{2a(a^2+1)}\geq0.$$ Done!