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Prove, $$(a_1+a_2+a_3+a_4+a_5+a_6)^2 \geq 6(a_1a_3+a_1a_5+a_3a_5+a_2a_4+a_4a_6+a_6a_2)$$

$\forall a_i\in \mathbb{R}$. Given that $ a_i>a_{i+1}$.

I tried this using Chebyshev's Inequality, $$6\left(\sum_{i=1}^6 a_i^2\right)\geq\left(\sum_{i=1}^6a_i\right)^2$$

But it doesn't help the conditions as the equality sign is the opposite. I then looked towards Cauchy-Schwarz but it didn't help much as I can use $ a_i>a_{i+1}$ there. Can someone help me solve this?

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2 Answers 2

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A nice solution has already been posted by Calvin Lin. As I said this one is based on the hint given by him in the first note.

Here it starts.

let,

$p=\sum_{i=1}^6 a_i$ ,

$q=a_1a_3+a_1a_5+a_3a_5+a_2a_4+a_4a_6+a_6a_2$

a

Assume,

$$f(x)=(x-a_1)(x-a_3)(x-a_5)+(x-a_2)(x-a_4)(x-a_6)$$

Claim:

$f(x)$ has all its roots real and distinct

Evidence of Claim:

Observe that $f(a_1)>0,f(a_2)<0,f(a_3)<0,f(a_4)>0,f(a_5)>0,f(a_6)<0$. So, there must be $3$ real intersects(roots) on the $x$-axis by $f(x)$.

So, now proceeding to the previous point.

The function is equivalent to

$$f(x)=2x^3-px^2+qx-r$$

And $$f'(x)=6x^2-2px+q$$

So, to have the roots of $f(x)$ to be real and distinct discriminant of $f'(x)$ should be positive.

i.e.

$p^2-6q>0$

Which is equivalent to the equality which to be proven.

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  • $\begingroup$ That's a very nice construction! I love how the inequality just falls out naturally, and it could be how the problem setters came up with it. $\endgroup$
    – Calvin Lin
    Mar 1, 2023 at 19:28
  • $\begingroup$ I am glad you liked it! :-) $\endgroup$
    – Leibniz-Z
    Mar 2, 2023 at 6:01
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Hint: To use (abuse) the condition, an often useful substitution is to set $$a_6 = a, a_5 = a+b,a_4 = a+b+c, \ldots , a_1 = a+b+c+d+e+f.$$ Lots of terms cancel out, in particular $a$ is no longer involved.
The rest of the variables are positive (resp non-negative if $ a_i \geq a_{i+1} $), so if the expression has no negative coefficients, then it's positive (resp non-negative).
As it turns out, there are negative coefficients, so how can we deal with them?

Notes:

  • If $a$ (which could be any real number) was involved, view it as a quadratic in $a$ and show that the discriminant is non-positive.
  • In fact, we have a strict inequality, given $a_i > a_{i+1}$.
  • Another start could be to define $ f(a_1, a_2, a_3, a_4, a_5, a_6) = (\sum a_i)^ 2 - 6(a_1a_3 + a_3a_5 + \ldots + a_6 a_2)$, and show that $f(a_1, a_2, a_3, a_4, a_5, a_6) = f(a_1-k, a_2-k, a_3-k, a_4-k, a_5-k, a_6-k)$. In particular, we can set $k = a_6$, which gets rid of 1 of the variables.
    • That this holds for all $k$ is equivalent to the above observation that $a$ is no longer involved.
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  • $\begingroup$ If I've made no arithmetic mistakes with the coefficients, this works, by rearranging to $(b-f)^2+4c^2+3d^2+4e^2+4bc+2be+6cd+4ce+2cf+6de+4ef\ge0$. $\endgroup$
    – J.G.
    Mar 1, 2023 at 16:35
  • $\begingroup$ Nice solution. I never thought this would work with replacements. And thanks to your hint in the notes 1. I've found an another way to solve this. But this was my question So I shouldn't post that answer. $\endgroup$
    – Leibniz-Z
    Mar 1, 2023 at 17:05
  • $\begingroup$ @Siddharth Please DO post an alternative solution! It's not disallowed (and I'd be excited to see your solution). $\endgroup$
    – Calvin Lin
    Mar 1, 2023 at 17:15

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