Let a,b,c be the sides of triangle such that $a+b+c=1$. Prove that $$5(ab+bc+ca)\geq18abc+a+b+c$$ I tried to prove: $$5(ab+bc+ca)\geq18abc+a+b+c$$$$10(ab+ac+bc)\geq36abc+2(a+b+c)$$$$a(5b+5c-2-12bc)+b(5c+5a-2-12ca)+c(5a+5b-2-12ab)\geq0.$$I tried to prove $$5b+5c-2-12bc\geq0.$$I know $$-bc\geq-\frac{(b+c)^2}{4}.$$I want to prove $$5(b+c)-2-3(b+c)^2\geq0$$$$5(1-a)-2-3(1-a)^2\geq0$$and I got $$a(1-3a)\geq0$$which can be wrong.

  • $\begingroup$ Your last inequality holds only when $0\le a\le \frac13$. But as $a$ is arbitrary; that is, it can be replaced by $b$ or $c$, and it is a side of a 'unit' triangle, doesn't this automatically hold? In a triangle, if $a+b+c=1$, then the smallest side has length of at most $\frac13$. $\endgroup$ – TheSimpliFire Aug 12 '18 at 13:55
  • $\begingroup$ In your proof, you multiply by 2 and collect terms in a manner that eludes me. My next 2 lines after the statement would be: $5ab+5bc+5ac-18abc-a-b-c≥0$ $a(5b+5c-18bc-1)+b(5c-1)-c$ $\endgroup$ – poetasis Aug 12 '18 at 14:11
  • $\begingroup$ It seems that you have expressed values of $c$ that have already been expressed in terms of $a$ and $b$, Likewise, a value of $b$ is already expressed in the parentheses of $a$. $\endgroup$ – poetasis Aug 12 '18 at 14:17
  • $\begingroup$ @José Carlos Santos I think this question has context (the topic starter thinks that we can use Schur for the proof) and also, he shows attempts. Why did you close this topic? $\endgroup$ – Michael Rozenberg Aug 14 '18 at 4:58

Let $a=y+z$, $b=x+z$ and $c=x+y$.

Thus, $x$, $y$ and $z$ are positives and we need to prove that: $$5(a+b+c)(ab+ac+bc)\geq18abc+(a+b+c)^3$$ or $$10(x+y+z)\sum_{cyc}(x^2+3xy)\geq8(x+y+z)^3+18\prod_{cyc}(x+y)$$ or $$\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0,$$ which is indeed Schur.


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