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}.

All Soviet Union Olympiad, 1987.

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. 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

    Like

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