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

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.

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

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: $$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+...+\frac{a_n^2}{b_n}\geq\frac{(a_1+a_2+...+a_n)^2}{b_1+b_2+...+b_n}.$$

• Thank you for your answer! Can you maybe tell me which $x_1,x_2,\dots$ you used in Cauchy-Schwarz?
– user708986
Oct 9, 2019 at 15:42
• @StackUnderflow I added something. See now. Oct 9, 2019 at 15:46
• Thank you! After Google search I found that this is known as Titu‘s Lemma
– user708986
Oct 9, 2019 at 17:25
• @StackUnderflow It's known as Cauchy-Schwarz inequality. This inequality was written in this form much more before than Titu Andreescu was born. Oct 9, 2019 at 17:34

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.