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Suppose $x_1x_2x_3=y^3$ and $x_1,x_2,x_3 \in \mathbb{R},x_1\neq x_2\neq x_3,\text{All of them are positive numbers}$
we need to show $$(1+x_1)(1+x_2)(1+x_3)\geq (1+y)^3$$ This inequality can proved by algebraic work ,but I am looking for a visual (or something like this) proof. I am thankful in advance for any idea .

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  • $\begingroup$ A question (not much effect on the main problem): If $x_1\neq x_2\neq x_3$, then when does the equality holds in the inequality? $\endgroup$ – MAN-MADE Sep 3 '17 at 17:22
  • $\begingroup$ There might be an inequality of this kind; but additional assumptions are needed. When $x_1=1$, $x_2=-2$, $x_3=-4$, $y=2$ then $x_1x_2x_3=y^3$, but your inequality is violated. $\endgroup$ – Christian Blatter Sep 3 '17 at 18:05
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Expanding both sides we get: $1+ x_1 + x_2+ x_1x_2 + x_3 + x_2x_3+x_1x_3+x_1x_2x_3>1+ 3y+3y^2+y^3$. Use AM-GM inequality next: $x_1+x_2+x_3 > 3y, x_1x_2+x_2x_3+x_3x_1 > 3y^2$, and add these up. Thus its proven.Note that this is still "algebraic" but it gets readers start. And in fact, the interpretation is just interpret geometrically the AM-GM inequality for $3$ positive numbers. Hope it helps. One of these popular topics could be found in Roger Nelsen's book on "Proofs Without Words" of the MAA back in the $90$'s.

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    $\begingroup$ FWIW the $n=2$ case of AM-GM has several simple "visual" proofs e.g. [1], [2], [3]. $\endgroup$ – dxiv Sep 3 '17 at 18:50

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