0
$\begingroup$

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.

$\endgroup$
0

3 Answers 3

Reset to default
2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ Thank you for your answer! Can you maybe tell me which $x_1,x_2,\dots$ you used in Cauchy-Schwarz? $\endgroup$
    – user708986
    Oct 9, 2019 at 15:42
  • 1
    $\begingroup$ @StackUnderflow I added something. See now. $\endgroup$
    – Michael Rozenberg
    Oct 9, 2019 at 15:46
  • $\begingroup$ Thank you! After Google search I found that this is known as Titu‘s Lemma $\endgroup$
    – user708986
    Oct 9, 2019 at 17:25
  • $\begingroup$ @StackUnderflow It's known as Cauchy-Schwarz inequality. This inequality was written in this form much more before than Titu Andreescu was born. $\endgroup$
    – Michael Rozenberg
    Oct 9, 2019 at 17:34
1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.