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 $(2a-1)^{2}$.

Answer 2

Since the vertex of the RHS is at $(0.5, 0.75)$, given taht the LHS when $a=1$ is $1$, if you can prove that the LHS is strictly increasing, you can complete the proof.

Hint: Simplify the LHS into a quadratic and find its vertex.

Answer 3

By C-S $$\frac{a^2}{a+1}+\frac{(1-a)^2}{2-a}\geq\frac{(a+1-a)^2}{a+1+2-a}=\frac{1}{3}.$$

I used the following C-S:

For any reals $a_i$ and positives $b_i$ we have: $$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+...+\frac{a_n^2}{b_n}\geq\frac{(a_1+a_2+...+a_n)^2}{b_1+b_2+...+b_n}.$$