Mathematical Reflections 2012, Issue 2 - Problem S224

Problem:
Let $a,b,c$ real numbers greater than $2$ such that $$\dfrac{7-2a}{3a-6}+\dfrac{7-2b}{3b-6}+\dfrac{7-2c}{3c-6}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}.$$ Prove that
$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \leq 1.$

Proposed by Titu Andreescu.

Solution:
Since $$\dfrac{7-2a}{3a-6}+\dfrac{7-2b}{3b-6}+\dfrac{7-2c}{3c-6}=-\dfrac{2}{3}+\dfrac{1}{a-2}-\dfrac{2}{3}+\dfrac{1}{b-2}-\dfrac{2}{3}+\dfrac{1}{c-2},$$ the given condition can be rewritten as $$\dfrac{1}{a(a-2)}+\dfrac{1}{b(b-2)}+\dfrac{1}{c(c-2)}=1.$$
Suppose without loss of generality that $a \leq b \leq c$. If $$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} > 1$$ then $\dfrac{1}{a-2}+\dfrac{1}{b-2}+\dfrac{1}{c-2} > 3$ and by Chebychev's Inequality,
$$3 < \left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(\dfrac{1}{a-2}+\dfrac{1}{b-2}+\dfrac{1}{c-2}\right) \leq 3,$$ contradiction.

Mathematical Reflections 2012, Issue 2 - Problem J227

Problem:
For a positive integer $N$ let $r(N)$ be the number obtained by reversing the digits of $N$.
Find all $3$-digit numbers $N$ such that $r^2(N)-N^2$ is the cube of a positive integer.

Proposed by Titu Andreescu.

Solution:
Let $N=100a+10b+c$ and $r(N)=100c+10b+c$ with $0 \leq a,b,c \leq 9$, $ac \neq 0$. Then
$$r^2(N)-N^2=(100c+10b+a)^2-(100a+10b+c)^2=99(c-a)[101(c+a)+20b].$$
Since $r^2(N)-N^2 > 0$, it must be $c > a$. If $r^2(N)-N^2$ is a cube, as $c-a \leq 8$, then $101(c+a)+20b \equiv 0 \pmod{121}$, i.e. $b-(a+c) \equiv 0 \pmod{121}$. Since $-17 \leq b-(a+c) \leq 6$, then $b=a+c$. So,
$$r^2(N)-N^2=11^3\cdot3^2(c-a)(c+a).$$
If $c-a=3$ then $c+a=2a+3$ and $5 \leq 2a+3 \leq 19$ must be an odd cube, contradiction. Likewise, $c-a=6$ yields $c+a=2(a+3)$, so that $a+3=2k^3$ where $k$ is a positive integer, contradiction. So, $c+a=3,6,9$.

(i) If $c+a=3$, then $a=1, c=2$ and it's easy to see that $r^2(N)-N^2=11^3\cdot3^3 = 33^3$.
(ii) If $c+a=6$, then $c-a=2(3-a)$ and $r^2(N)-N^2=11^3\cdot3^3\cdot2^2(3-a)$, so $a=1$ and $c=5$.
(iii) If $c+a=9$, then $c-a=9-2a$ and $r^2(N)-N^2=11^3\cdot3^4 (9-2a)$, but $9-2a=9k^3$  has no positive integer solution.

In conclusion, all the $3$-digit numbers which satisfy the required conditions are $132$ and $165$.

Mathematical Reflections 2012, Issue 2 - Problem J225

Problem:
Let $a, b, c$ be nonnegative real numbers such that $a+b+c=1$. Prove that $$ab(a^2+b^2)+bc(b^2+c^2)+ca(c^2+a^2)+abc \leq \dfrac{1}{8}.$$

Proposed by Titu Andreescu.

Solution:

Using the fact that $abc=abc(a+b+c)$ and factorizing we have
$$ab(a^2+b^2)+bc(b^2+c^2)+ca(c^2+a^2)+abc=(a^2+b^2+c^2)(ab+bc+ca).$$
By AM-GM Inequality, $$a^2+b^2+c^2+2(ab+bc+ca)=1 \geq 2\sqrt{2(a^2+b^2+c^2)(ab+bc+ca)},$$
then $$(a^2+b^2+c^2)(ab+bc+ca) \leq \dfrac{1}{8}.$$  

Mathematical Reflections 2012, Issue 2 - Problem J223

Problem:
Let $a$ and $b$ be real numbers such that $\sin^3 a - \dfrac{4}{3} \cos^3 a \leq b-\dfrac{1}{4}$. Prove that
$$\dfrac{3}{4}\sin a - \cos a \leq b+\dfrac{1}{6}.$$

Proposed by Titu Andreescu.

Solution:
We argue by contradiction. Suppose that $b+\dfrac{1}{6} < \dfrac{3}{4}\sin a - \cos a$. Then, summing up the two inequalities and rearranging, we have $$\dfrac{5}{12} < \dfrac{1}{4} \sin 3a + \dfrac{1}{3} \cos 3a.$$
But $$\dfrac{1}{4} \sin 3a + \dfrac{1}{3} \cos 3a = \dfrac{5}{12} \left(\dfrac{4}{5}\cos 3a + \dfrac{3}{5} \sin 3a\right) = \dfrac{5}{12} \cos \left(3a-\arctan \dfrac{3}{4}\right) \leq \dfrac{5}{12},$$ a contradiction.