Show that if $a,b$ and $c,d$ are positive (and $c$ and $d$ are rational), then $(a^c-b^c)(a^d-b^d)\geq$0 and $a^{c+d}+b^{c+d}\geq a^c b^d+a^db^c$. Under what circumstance do the signs of equality hold?
My attempt,
I tried to expand the expression $(a^c-b^c)(a^d-b^d)=(a^{c+d}-a^cb^d-b^ca^d+b^{c+d})$. But it seems this expression doesn't mean anything, maybe it does, but I've no idea how. Could someone give me some hints? Thanks a lot.
On
The expanded expression gives the second inequality if the first one hold : $$a^{c+d}-a^cb^d-a^db^c+b^{c+d}\ge0\iff a^{c+d}+b^{c+d}\ge a^cb^d+a^db^c$$ The first one holds from the fact that $t\mapsto t^x$ is increasing for $x>0$, so $a^c-b^c$ and $a^d-b^d$ are both positive or negative together with $a-b$.
For positives $c$ and $d$ we obtain:
Let $a\geq b$.
Thus, $a^c\geq b^c$ and $a^d\geq b^d$, which says $(a^c-b^c)(a^d-b^d)\geq0$.
Let $a\leq b$.
Thus, $a^c\leq b^c$ and $a^d\leq b^d$, which says $(a^c-b^c)(a^d-b^d)\geq0$.
Done!