77. Given that $x$ and $y$ are positive real numbers, let $M=\frac{x}{2 x+y}+\frac{y}{x+2 y}$, $N=\frac{x}{x+2 y}+\frac{y}{2 x+y}$. If there exists a constant $t$ such that the maximum value of $M$ is $t$, and the minimum value of $N$ is also $t$,
(1) Guess the value of $t$,
(2) Prove that your guess is correct.
Solution: (1) Let $\mathrm{x}=1, \mathrm{y}=1$, then $M=\frac{2}{3}$ and $N=\frac{2}{3}$, conjecture $t=\frac{2}{3}$
(2) (1) First prove that $M=\frac{x}{2 x+y}+\frac{y}{x+2 y}$ has a maximum value of $\frac{2}{3}$

Because $\frac{x}{2 x+y}+\frac{y}{x+2 y}-\frac{2}{3}$
$$
\begin{array}{l}
=3 x(x+2 y)+3 y(2 x+y)-2(2 x+y)(x+2 y) \\
=3 x^{2}+12 x y+3 y^{2}-\left(4 x^{2}+10 x y+4 y^{2}\right) \\
=-x^{2}+2 x y-y^{2} \\
=-(x-y)^{2} \leq 0
\end{array}
$$

When and only when $x=y$, the equality holds.
So when $x=y>0>0$, $M=\frac{x}{2 x+y}+\frac{y}{x+2 y}$ has a maximum value $t=\frac{2}{3}$
(2) Then prove: $N=\frac{x}{x+2 y}+\frac{y}{2 x+y}$ has a minimum value of $\frac{2}{3}$
Because $\frac{x}{x+2 y}+\frac{y}{2 x+y}-\frac{2}{3}$
$$
\begin{array}{l}
=3 x(2 x+y)+3 y(x+2 y)-2(x+2 y)(2 x+y) \\
=6 x^{2}+6 x y+6 y^{2}-\left(4 x^{2}+10 x y+4 y^{2}\right) \\
=2 x^{2}-4 x y+2 y^{2} \\
=2(x-y)^{2} \geq 0
\end{array}
$$

When and only when $x=y$, the equality holds.
So when $x=y>0>0$, $N=\frac{x}{x+2 y}+\frac{y}{2 x+y}$ has a minimum value $t=\frac{2}{3}$