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Let $ x,y,z \in \mathbb{R_{+}^{*}} $ , with $ xy+yz+zx=1 $.

Prove that: $$4(x+y+z)\leq 9xyz+\frac{1}{xyz}$$

I tried to solve the inequality with the Cardano-Tartaglia formula but I havent't succeeded so far.

Xam
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ztefelina
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    OK, a boring approach. Original inequality $\iff 4(x+y+z)xyz\leq 9x^2y^2z^2+1 \iff 4(x+y+z)xyz(xy+yz+zx)\leq 9x^2y^2z^2+(xy+yz+zx)^3$

    For the last one, RHS-LHS $\displaystyle= \sum_{sym}x^3y^3+3x^2y^2z^2-xyz\sum_{sym}x^2y$. Then you can try to use things like this or this

     –  Dec 12 '16 at 22:19
  • Let $t^3+bt^2+ct+d$ be the cubic whose roots (with multiplicity) are $x,y,z$. Then you have $c=xy+yz+zx=1$, and the condition you want to prove is $$ 4\cdot(-b) \le 9\cdot(-d) + \frac1{-d} $$ which is the same as $$ 4bd \ge 9d^2 + 1 $$ From there I would expect you can use the discriminant of the cubic (which must be non-negative) for something. – hmakholm left over Monica Dec 12 '16 at 22:23

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Let $a=xy$, $b=yz$, and $c=zx$. Multiply both sides of the desired inequality by $xyz$: we are to prove $$ 4(x+y+z)xyz\leq 9(xyz)^2+1\iff 4(ac+ab+bc)\leq 9(abc)+1 $$ whenever $a,b,c>0$ satisfy $a+b+c=1$. Rewrite this once more as $$ 4(ac+ab+bc)(a+b+c)\leq9abc+(a+b+c)^3 $$ which you can expand both sides and then do some cancelling to get $$ ab(a+b)+ac(a+c)+bc(b+c)\leq 3abc+a^3+b^3+c^3\tag{$*$}. $$ But ($*$) is a consequence of the Schur's inequality so it holds and so does the original inequality.

yurnero
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