TITLE: How to prove this inequality with the simplest means?
QUESTION [2 upvotes]: $$ x^2+5y^2+6z^2 \geq 4\sqrt{5xyz^2},\ \text{ if }\ xy>0 $$
I was trying to prove it. The right hand side should be nonnegative, so I can square it on both sides.
But once I have done it I get to a point, where I do not see how to show, that the statement is greater or equal to $0.$
$$x^4+25y^4+36z^4+10x^2y^2+12x^2z^2+60y^2z^2-80xyz^2 \geq 0. $$
Thank you

REPLY [4 votes]: Looks like the AM.GM inequality will by useful.  We might as well assume $x,y>0$. (As $xy>0$ then either $x,y > 0$ or $x,y < 0$.  If $x,y < 0$ then we can replace $x,y$ with $|x|, |y|$ and the result will be the same.... Or... I'll just use $|x|,|y|$).
By AM.GM we have $x^2 + 5y^2 \ge 2\sqrt{x^2\cdot 5y}=2|x||y|\sqrt 5= 2xy\sqrt 5$.
So $x^2 + 5y^2 + 6z^2 \ge 2xy\sqrt 5 + 6z^2$.
Apply AM.GM on $2xy\sqrt 5$ and $6z^2$ and we have $2xy\sqrt 5 +6z^2 \ge 2\sqrt{2xy\sqrt 5\cdot 6z^2}=2\sqrt{12xyz^2\sqrt 5}=4\sqrt{3\sqrt 5xyz^2}$.
And as $3 > \sqrt 5$ we have $4\sqrt{3\sqrt 5xyz^2}>4\sqrt{\sqrt 5^2xyz^2}=4\sqrt{5xyz^2}$
And that does it.
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