Backwards induction to show that $x_1\cdots x_n \leq ((x_1+\cdots+x_n)/n )^n$

My question is about backwards induction and was asked previously in the above link. How did they derive the equations for $n=2$, which is $(x_1+x_2)^2−4x_1x_2=(x_1−x_2)^2≥0$

Here's what I have for $n=2$:

$x_1x_2\leq((x_1+x_2)/2)^2$ =$x_1x_2\leq(x_1+x_2)^2/4$

And that's about as far as I get. Can somebody give me any pointers?

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A surprising number of inequalities come down, sooner or later, to the fact that any square is non-negative. –  André Nicolas Dec 13 '12 at 4:39

Your inequality, $x_1x_2\le\frac14(x_1+x_2)^2$, is equivalent to the inequality $4x_1x_2\le(x_1+x_2)^2$: just multiply through by $4$. This in turn is equivalent to $(x_1+x_2)^2-4x_1x_2\ge 0$: just subtract $4x_1x_2$ from both sides. Now expand the square to see that the inequality can also be written

$$x_1^2+2x_1x_2+x_2^2-4x_1x_2\ge 0\;,$$

do the algebra on the lefthand side to get

$$x_1^2-2x_1x_2+x_2^2\ge 0\;,$$

and factor the familiar square to write it finally as $(x_1-x_2)^2\ge 0$. This inequality is obviously true, and you can reverse all of the algebraic steps above to see that it’s equivalent to the one from which you started.

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Thanks so much! –  dylam Dec 13 '12 at 4:29
@yoyoyo: You’re welcome! –  Brian M. Scott Dec 13 '12 at 4:31