Prove $f(x,y) > g(x,y)$ for all $x,y \in [0,1]$ I'm trying to prove the following:
$$ 4xy + 4(1-x)(1-y) < \max\{8xy,8(1-x)(1-y),3\}  \qquad \forall x,y \in [0,1] $$
In the language from the class, I'm trying to show that: $m_2 < \max\{m_1,m_3,m_4\}$
I think that's written correctly. It's derived from a microeconomics game theory problem with $3$ players. I'm trying to prove that some choice ($m_2$) has an expected value that is strictly less than at least one of the three other choices ($m_1, m_3, m_4$) no matter how $x$ and $y$ vary, assuming $x$ and $y$ between $0$ and $1$ inclusive. ($x$ and $y$ are probability variables.)
Spot checks show that this is the case, but I assume calculus can be used to prove it for all cases. I don't know if it's required of us, but I want to know nonetheless.


*

*for $x=0, y=0$: $m_2 < m_3$. 

*for $x=1, y=1$: $m_2 < m_1$. 

*for $x =.5, y=.5$: $m_2 < m_4$. 
and so on. 
My intuition says the answer somehow involves doing something with "level set" but that's getting to the furthest reaches of my math knowledge. 
Thanks, I hope I'm asking this correctly!

Say that $a=4xy$ and $b=4(1-x)(1-y)$. If $a<b$, we know that $a+b<b+b=2b=8(1-x)(1-y)$. If $a>b$, we know that $a+b<a+a=2a=8xy$. If $a=b$, we have $xy=(1-x)(1-y)$. We then want to show that $a+b<3$, since $a+b=2a=2b$. Equality holds if and only if $x=1-y$. Then, the maximum of 
$$
4xy+4(1-x)(1-y)=8y(1-y)
$$
occurs at $y=\frac 12$. The maximum is $8\cdot \frac 12\cdot \frac 12=2<3$. Thus, we always can find a part of the right hand side that is strictly larger than the left hand side, $a+b$, and therefore the inequality holds for all $x,y\in[0,1]$.
