# 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

## 4 thoughts on “A gem of an inequality from USSR !!!”

1. seriouscephalopod

Neat. I’m seeing an alternate solution where you start with the identity describing the inequality of arithmetic and geometric means $\sqrt{xy} \leq (x + y)/2$ in the equivalent form

$xy \leq (x + y)^2 / 4$

which applied to $aA, bB, cC$ yields $aA \leq (a + A)^2/4 = k^2 / 4$ $bB \leq (b + B)^2/4 = k^2 / 4$ $cC \leq (c + C)^2/4 = k^2 / 4$

and then summing these we get the slightly stronger inequality $aA + bB + cC \leq 3k^2 / 4$

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• nkpithwa

Thanks for participating !!

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2. seriouscephalopod

Oh, and there, 3s later I see I misread the identity…Nevermind.

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• nkpithwa

Yes, many thanks.

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