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Let $0\leq a\leq b\leq c\leq d\leq e\leq f$. Prove that $$a^2-f^2+abd-acd+ace-abe-bcf+cdf+bef-def\leq 0.$$

I have tried to split it into pairs and prove that each such pair is negative. However, I have been unsuccessful with this approach as of now.

This of course is equivalent to the following inequality, which might be easier to see than the previous one: $$\frac{a^2-f^2+abd-acd+ace-abe-bcf+cdf+bef-def}{a-f}\geq 0.$$

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  • $\begingroup$ Could there be a typo? $+ace-abe$ might be $-ace+abe$ me think. $\endgroup$
    – m-stgt
    Mar 21 at 14:01
  • $\begingroup$ In addition, your 2nd ansatz will badly fail if $a=f$ what is possible by $0\leq a\leq b\leq c\leq d\leq e\leq f$. $\endgroup$
    – m-stgt
    Mar 21 at 14:05

2 Answers 2

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Write the inequality in the form $(a+f)(a-f)+a(c-b)(e-d)-f(d-b)(e-c) \leq 0$. It is then clear that $(a+f)(a-f) \leq 0$ and $\begin{align*} a(c-b)(e-d)-f(d-b)(e-c) &\leq f(c-b)(e-d)-f(d-b)(e-c) \\ &\leq f((d-b)(e-c)-(d-b)(e-c))=0. \end{align*}$

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I would try to split the whole thing in pairs:

You know that $a \le f$, so $a^2 \le f^2$, so $a^2-f^2 \le 0$.
Also you know that $b \le c$, hence $ad \cdot b \le ad \cdot c$, meaning that $abd \le acd$ or $abd - acd \le 0$.

Try to continue like this, creating all such pairs.

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