可以通过在定义友元函数时添加friend bool operator<(const MyClass& a, const MyClass& b) __attribute__((always_inline));的方式来解决此问题。其中，__attribute__((always_inline))用于告诉编译器一定要内联该函数。以下是示例代码：

#include 

using namespace std;

class MyClass {
    int a, b;
public:
    MyClass(int i, int j) { a = i; b = j; }
    friend bool operator<(const MyClass& a, const MyClass& b) __attribute__((always_inline));
};

bool operator<(const MyClass& a, const MyClass& b) {
    return (a.a + a.b) < (b.a + b.b);
}

int main() {
    MyClass obj1(2, 3), obj2(1, 5);
    if (obj1 < obj2) {
        cout << "obj1 is less than obj2\n";
    }
    else {
        cout << "obj2 is less than obj1\n";
    }
    return 0;
}