# Prove that $a^2-f^2+abd-acd+ace-abe-bcf+cdf+bef-def\leq 0.$ [closed]

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: $$\frac{a^2-f^2+abd-acd+ace-abe-bcf+cdf+bef-def}{a-f}\geq 0.$$

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

## 2 Answers

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*}

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.