# Proving an inequality with $6$ variables and a constant $C$

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?

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

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.