question about AM-GM inequality ab<(a^2+b^2)/2

Question

I have a question about the AM-GM inequality if a,b>0 then

1. ab≤(a+b)^2/4;
2. ab≤a^2+b^2;

I wonder if these 2 inequality are true for all a,b>0, if not which one is correct?

Answer 1

Note that by AM-GM inequality, if $c,d>0$ then ; $$\sqrt{cd}\leq \frac{c+d}{2}$$ Substitute $c = a^2 ; d = b^2$ for $a,b>0$ to get $$ab\leq \frac{a^2+b^2}{2} $$ $$\implies 2ab\leq a^2+b^2$$ $$\implies 4ab \leq a^2+b^2+2ab = (a+b)^2$$ $$ \implies ab\leq \frac{(a+b)^2}{4}$$ Hence first inequality is true.
Now note the second equation that is $$ab\leq \frac{a^2+b^2}{2} $$ But $\frac{a^2+b^2}{2}<a^2+b^2$ this implies $$ab< a^2+b^2$$ Hence the second inequality in your question is also true but without the equality sign.