# A gem of an inequality from USSR !!!

Problem:

The positive numbers a, b, c, A, B, C satisfy

$a+A=b+B=c+C=k$.

Prove that $aB+bC+cA \leq k^{2}$.

Solution:

Proving that $aB+bC+cA \leq k^{2}$ is equivalent to showing that

$aB+bC+cA = a(k-b)+b(k-c)+c(k-a) \leq k^{2}$

$\Longleftrightarrow k(a+b+c) \leq k^{2}+ab+bc+ca$

But this is equivalent to

$k^{2}-ka+bc \geq (b+c)(k-a) \Longleftrightarrow bc \geq (k-a)(b+c-k)$

If $b+c-k \leq 0$, then the above is obviously true. If not, then we have

$(k-b)(k-c) \geq 0 \Rightarrow bc \geq k(b+c-k) \geq (k-a)(b+c-k)$

and thus, our inequality holds.

More later,

Nalin Pithwa

(Russia 2001).

Let a and b be distinct positive integers such that $ab(a+b)$ is divisible by $a^{2}+ab+b^{2}$. Prove that

$|a-b|> \sqrt[3]{ab}$.

Proof.

Let $g=gcd(a,b)$ and write $a=xg$ and $b=yg$ with $gcd(x,y)=1$. Then,

$\frac{ab(a+b)}{a^{2}+ab+b^{2}}=\frac{xy(x+y)g}{x^{2}+xy+y^{2}}$

is an integer. N^ote that $gcd(x^{2}+xy+y^{2}, x)=gcd(y^{2},x)=1$. Similarly, $gcd(x^{2}+xy+y^{2},y)=1$.

Because $gcd(x+y,y)=1$, we have

$gcd(x^{2}+xy+y^{2},x+y)=gcd(y^{2}, x+y)=1$

Now, we apply the following lemma: Let a and b be two coprime numbers. If c is an integer such that $a|c$, then $ab|c$.

Hence, we get, $x^{2}+xy+y^{2}|g$ implying that $g \geq x^{2}+xy+y^{2}$. Therefore,

$|a-b|^{3}=|g(x-y)|^{3}=g^{2}|x-y|^{3}.g \geq g^{2}.1.(x^{2}+xy+y^{2})>g^{2}xy=ab$.

It follows that $|a-b|>\sqrt[3]{ab}$. QED.

Note that the key step $x^{2}+xy+y^{2}$ divides g can also be obtained by clever algebraic manipulations such as $a^{3}=(a^{2}+ab+b^{2})a-ab(a+b)$.