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Find the greatest constant $C$ so that for all real numbers $x_1, x_2,\ldots,x_6$ the inequality $$(x_1 + x_2 + ...+ x_6)^2 \geq C \cdot (x_1 (x_2 + x_3) + x_2 (x_3 + x_4) + \ldots+ x_6 (x_1 + x_2))$$ applies.

Can someone give me a hint as to how to solve this?

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  • $\begingroup$ Have you tried expanding both sides and consolidating terms? $\endgroup$
    – Rushabh Mehta
    Mar 17, 2021 at 20:55
  • $\begingroup$ $C$ should be equal to $3$. $\endgroup$
    – NN2
    Mar 17, 2021 at 21:01

1 Answer 1

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Hint: It is well known that $(a+b+c)^2 \geq 3 (ab+bc+ca)$, with equality when $ a = b = c$.

Further hint, (actually the crux in this approach, so I encourage you to think about how the previous line could be a hint. If you don't know what to do, I suggest to skip this part first and read the rest of the solution below the separator):

Set $ a = x_1 + x_4, b = x_2 + x_5, c = x_3 + x_6$.

Hence $ C = 3$.

Steps that one could take to approach the inequality directly:

  1. Setting $ x_i = 1$ gives $ C \leq 3$. A reasonable guess is that $ C = 3$ is the answer (though this need not be the case).

  2. Notice that with $C=3$, equality holds when $ x_1 = x_3 = x_5 , x_2 = x_4 = x_6$, so it's not a standard "all terms equal" scenario.

  3. Calling the term on the RHS $X$ (without the constant), show that $$X = (x_1 + x_4) ( x_2 + x_5) + (x_2 + x_5) ( x_3 + x_6) + (x_3 + x_6) ( x_1 + x_4).$$ Arguably, this is a "nicer" expression.

  4. Show that
    $$(x_1 + x_2 + x_3 + x_4+ x_5 + x_6) ^2 - 2X = (\sum x_i^2) + 2 ( x_1x_4 + x_2x_5 + x_3x_6) \geq X.$$
    Equality holds when $x_1 + x_4 = x_2 + x_5 = x_3 + x_6$.

  5. Hence $(x_1 + x_2 + x_3 + x_4+ x_5 + x_6) ^2 \geq 3X$

Hence, the answer is $C = 3$.

From here, it's easy to motivate the original substitution.

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