Inequality $\frac{a^2}{a+1}+\frac{(1-a)^2}{2-a}\geq\frac13$

Question

Let $0\le a\le 1$. How can I prove that $$\frac{a^2}{a+1}+\frac{(1-a)^2}{2-a}\geq\frac13$$ ?

I tried multiplying everything by $(a+1)(2-a)$ which leaves me with
$$a^2(2-a)+(1+a)(1-a)^2\geq \frac{(a+1)(2-a)}{3}$$ but using a.m.-g.m. on the left-hand-side didn’t work for me.

Answer 1

The inequality holds if and only if $3a^{2}(2-a)+3(1-a)^{2}(a+1)-(a+1)(2-a)\geq 0$, the former is just $4a^{2}-4a+1$ which is just $(2a-1)^{2}$.