Friday, May 27, 2016

Prove that $x^2 + y^2+ z^2\le xyz + 2$ where the reals $x,\,y,\, z\in [0,1]$.

Prove that $x^2 + y^2+ z^2\le xyz + 2$ where the reals $x,\,y,\, z\in [0,1]$.

For all $x,\,y,\, z\in [0,1]$, we know $x^2 + y^2+ z^2\le x+y+z$.

And if we can prove $x+y+z\le xyz + 2$, we're done. Note that from the conditions $x≤1$ and $y≤1, z≥0$, we can set up the inequality as follows:

$(1-x)(1-y)(z)≥0$, upon expanding we get $xyz≥z(x+y)-z$, adding a 2 on both sides yields $2+xyz≥z(x+y)-z+2$ and it's trivial in proving $z(x+y)-z+2≥x+y+z$ holds for $x,\,y,\, z\in [0,1]$ since:

$z(x+y)-z+2≥x+y+z$

$z(x+y)≥x+y+2z-2$

$(x+y)(z-1)≥2(z-1)$

We know from the given conditions that:

1.
$z≤1$, this tells us $z-1≤0$,

2.
$x≤1,\,y\le 1$ so adding them up yields $x+y\le 2$,

Therefore if we multiply the quantity of $z-1$ to both sides of $x+y\le 2$, we have to reverse the inequality sign and the result is then follows.