Inequality Challenge: Prove $x^2+y^2+z^2\le xyz+2$ [0,1]

[anemone]

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

 

[kaliprasad]

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

Spoiler

we have $x^2+y^2+z^2-xyz = x(x-yz) + y^2 + z^2$
keeping y and z fixed this increases when x increased
similarly keeping x and z fixed this increases when y increases and keeping x and y fixed this increases when z increases.
so this increases when x,y,z increase and maximum value is when x =y=z = 1( that is the range)
so $x^2+y^2+z^2 - xyz <= 1 +1 + 1 -1$
or $x^2+y^2+z^2 - xyz <= 2$
or $x^2+y^2+z^2 <= 2+xyz$

 

[anemone]

Spoiler

we have $x^2+y^2+z^2-xyz = x(x-yz) + y^2 + z^2$
keeping y and z fixed this increases when x increased
similarly keeping x and z fixed this increases when y increases and keeping x and y fixed this increases when z increases.
so this increases when x,y,z increase and maximum value is when x =y=z = 1( that is the range)
so $x^2+y^2+z^2 - xyz <= 1 +1 + 1 -1$
or $x^2+y^2+z^2 - xyz <= 2$
or $x^2+y^2+z^2 <= 2+xyz$

Good job kaliprasad!

My solution:

Spoiler

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 $(x+y)(z-1)≥2(z-1)$ is true, so the result follows.