It is correct, but I agree the reasoning is not written out properly. I’ll write y for y∗
We are in the case g(x)>x2. So if g(y)>=y2, then g(x)+g(y)>x2+y2>=2xy, where the last inequality is true because it’s equivalent to (x−y)2>=0. So we have g(x)+g(y)>2xy, which is a contradiction to the choice of y.
It is correct, but I agree the reasoning is not written out properly. I’ll write y for y∗
We are in the case g(x)>x2. So if g(y)>=y2, then g(x)+g(y)>x2+y2>=2xy, where the last inequality is true because it’s equivalent to (x−y)2>=0. So we have g(x)+g(y)>2xy, which is a contradiction to the choice of y.