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 .

  • $\begingroup$ 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

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.

| cite | improve this answer | |
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.