Solutions to two algebra problems for RMO practice

Problem 1.

If a, b, c are non-negative real numbers such that $(1+a)(1+b)(1+c)=8$ , then prove that the product abc cannot exceed 1.

Solution I:

Given that $a \geq 0$ , $b \geq 0$ , $c \geq 0$ , so certainly $abc>0$ , $ab>0$ , $bc>0$ , and $ac>0$ .

Now, $(1+a)(1+b) = 1 + a + b + ab$ and hence, $(1+a)(1+b)(1+c) = (1+a+b+ab)(1+c)= 1+a+b+ab+c +ac + bc + abc=8$ , hence we get:

$a+b+c+ab+bc+ca+abc=7$ . Clearly, the presence of $a+b+c$ and $abc$ reminds us of the AM-GM inequality.

Here it is $AM \geq GM$ .

So, $\frac{a+b+c}{3} \geq (abc)^{1/3}$ .

Also, we can say: $\frac{ab+bc+ca}{3} \geq (ab.bc.ca)^{1/3}$ . Now, let $x=(abc)^{1/3}$ .

So, $8 \geq 1+3x+3x^{2}+x^{3}$

that is, $8 \geq (1+x)^{3}$ , or $2 \geq 1+x$ , that is, $x \leq 1$ . So, this is a beautiful application of arithmetic mean-geometric mean inequality twice. 🙂 🙂

Problem 2:

If a, b, c are three rational numbers, then prove that : $\frac{1}{(a-b)^{2}} + \frac{1}{(b-c)^{2}} + \frac{1}{(c-a)^{2}}$ is always the square of a rational number.

Solution 2:

Let $x=\frac{1}{a-b}$ , $y=\frac{1}{b-c}$ , $z=\frac{1}{c-a}$ . It can be very easily shown that $\frac{1}{x}+ \frac{1}{y} + \frac{1}{z} =0$ , or $xy+yz+zx=0$ . So, the given expression $x^{2}+y^{2}+z^{2}=(x+y+z)^{2}$ is a perfect square !!! BINGO! 🙂 🙂 🙂