Could someone please help me with this AM-GM inequality question? I've sort of worked out the first bit, but I don't know how to solve the equality.

Question

sepeario · Answer

$$x^2 y^2 + x^2 + y^2 +4 \ge 6xy$$ The first section is to show that for all real x, y that the inequality holds. Then, I must solve for what values of x and y the EQUALITY holds.

legomyego180 · Answer

Interesting question. Id start by getting your numbers on one side of the inequality and your variable on the other. Then I'd factor the numerator.

sepeario · Answer

You mean like moving the 4 onto the right hand side, and the 6xy onto the left?

sepeario · Answer

I was able to prove that x is greater than y from using $$x^2 + y^2 \ge 2 \sqrt{x^2 y^2 }$$ and then factorising.

kainui · Answer

Hmm well one way to show equality with inequalities is if you know:

$$A \le B$$

and you can somehow perhaps use the AM-GM inequality to show:

$$B \le A$$

then you've shown that 

$$A=B$$

sepeario · Answer

Nevermind lol I don't actually know how to prove the inequality I stuffed up with my maths... @Kainui

sepeario · Answer

Would you know which part of the equation I would apply to the AM-GM inequality?

bobo-i-bo · Answer

I've worked it out. The key thing to answer the second bit is that the AM-GM inequality is an equality ONLY WHEN all the numbers are equal. In this case, there are only $x$ and $y$, so the AM-GM inequality is an equality only when $x=y$.

bobo-i-bo · Answer

Also, please see how you solved the first bit so that I know you haven't barked up the wrong tree?

raden · Answer

x^2y^2 + x^2 + y^2 + 4 >= 6xy
x^2y^2 + x^2 + y^2 + 4 - 6xy >= 0
(xy - 2)^2 + (x - y)^2  >= 0
the square number cant be a negative number, so we are done. the equality hold on when x = y and xy = 2. so, we get x = y = sqrt(2)