Csdrhrt

I don't understand why does the $theta (1-theta)(f(y)-f(x))(g(x)-g(y)) le 0 $ means both f and g are increasing or decreasing,can anyone take an example?

By the way,the solution said that $theta (1-theta)(f(y)-f(x))(g(x)-g(y)) le 0 $ means both f and g are increasing or decreasing,however,in the last formula,it just let $theta (1-theta)(f(y)-f(x))(g(x)-g(y)) = 0 $,why? Is't it $le 0$,not just $=0$ ?

Your interpretation of the logic is not quite right. It is saying that IF $f$ and $g$ are both increasing or both decreasing, THEN the third term is non-positive. Your interpretation is the converse statement: IF the third term is non-positive, THEN $f$ and $g$ are both increasing or both decreasing. In general, a statement is not logically equivalent to its converse.

Notice then that the correct interpretation certainly holds: If $f$ and $g$ are both increasing or both dcreasing, then $f(y)-f(x)$ and $g(x)-g(y)$ will have different signs, and since $0 le theta le 1$, we have $theta(1-theta) ge 0$. Hence
$$theta(1-theta)(f(y)-f(x))(g(x)-g(y)) le 0.$$

As for the final conclusion, I'm going to assume that we are given information in the problem statement that allows us to say $f$ and $g$ are both increasing or both decreasing. Then by the previous remarks, we have an inequality of the form
$$f(theta x + (1-theta)y)g(theta x + (1-theta)y) le A + B + C,$$
where $C le 0$. In the last step we aren't claiming $C = 0$, we are simply observing that adding nothing gives a bigger number than adding a non-positive number, so if we throw out the $C$ we still have an inequality.