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Let $a,b,c>0$ then we have : $$(5832ab+5832ac-216a+5832bc-216b-216c+6)^2\geq 4(54a+54b+54c-3) (472392abc-8748ab-8748ac+162a-8748bc+162b+162c-3)$$

As Wolfram alpha says there is a global minimum .

I try to expand but it gives nothing .

I can solve special case:

when $c=\frac{1}{54}$ it gives :

$$4 (1 - 54 a)^2 (1 - 54 b)^2\geq 0$$

When $a=b$ it gives :

$$11664 (1 - 54 b)^2 (b - c)^2\geq 0$$

Maybe Buffalo's way can solve it .

Thanks a lot for your time and patience .

Michael Rozenberg
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Miss and Mister cassoulet char
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1 Answers1

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Let $a=\frac{x+1}{54},$ $b=\frac{y+1}{54}$ and $c=\frac{z+1}{54}.$

Thus, we need to prove that: $$(xy+xz+yz)^2\geq3(x+y+z)xyz$$ or $$\sum_{cyc}z^2(x-y)^2\geq0,$$ which says that our inequality is true even for any reals $a$, $b$ and $c.$

Michael Rozenberg
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  • Your solution is very nice and beautiful, +1! But I have a question, how did you find this substitution? I saw it very hard for me and I really hope for your answer. – NKellira Feb 17 '20 at 11:59
  • @tthnew I saw a topic-starter's attempts and tried to get $xyz-xy-xz-yz+x+y+z-1=(x-1)(y-1)(z-1)$ in the RHS and it turned out. After this easy to get the final's substitutions. – Michael Rozenberg Feb 17 '20 at 12:02