# Tag Info

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When $0\le a,b,c \le 1$ or $a,b,c \ge 1$, we have $$(1-a)(1-bc) \ge 0,~~ (1-b)(1-ca) \ge 0,~~ (1-c)(1-ab) \ge 0~~~~(1).$$ Expanding and adding these three results and next by adding $1+abc$ on both sides, we prove that $$(1+a)(1+b)(1+c) \le 4+4 abc~~~~(2).$$ For other positive numbers, it (2) may or may not hold.

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It is obvious without even doing any arithmetic that if $z=1000$, there is nothing that $x$ or $y$ could possibly be to overcome the huge number that is $z^2$. The LHS grows in $O(n^2)$ and the RHS grows only in $O(n)$. That tells you that there is a computable upper limit to how large the (positive) values of $x,y,z$ can be. The question can be reduced ...

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I suppose you need a slightly modified version of the inequalities. Recall that all $x,y,z$ are integers, and in fact, we can think of them as of naturals. Then, for any $x,y,z \in \mathbb{N}$ and $y \neq 2$, $z \neq 2$, $z\neq 3$, we have $$x+\frac{2}{x}\geq 3,\quad y+\frac{3}{y} \geq 4, \quad z+\frac{4}{z} \geq 5.$$ Therefore, if $y\neq 2$, $z \neq 2$ ...

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Note that $$\frac{1}{b^3+1} - \Big(1 - \frac{b^2+b}{4}\Big) = \frac{b(b^2+3b+1)(b-1)^2}{4(b+1)(b^2-b+1)}\ge 0.$$ It suffices to prove that $$\sum_{\mathrm{cyc}} a^2\Big(1 - \frac{b^2+b}{4}\Big) \ge \frac{3}{2}$$ or $$a^2+b^2+c^2 - \frac{1}{4}(a^2b^2+b^2c^2+c^2a^2) - \frac{1}{4}(a^2b+b^2c+c^2a)\ge \frac{3}{2}.$$ Note that a^2b+b^2c+c^2a \le \sqrt{(a^2b^2+b^...

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