Triangle inequality equality case in integrals Consider the inequality
$$\left(\left(\int{f}\right)^{p}+\left(\int{g}\right)^{p}\right)^{1/p} \le \int{(f^p+g^p)^{1/p}}$$ 
where $f$ and $g$ are nonnegative measurable functions and $p \ge 1$ .
I believe that I can show it's true by MCT and simple functions, but what is its equality case?
My guess is that f and g are multiples up to a null set, but I'm not sure how to prove this.
 A: Take $f=F^{1/p}$ and $g=G^{1/p}$. Then you get
$$\left(\int F^{1/p}\right)^p+\left(\int G^{1/p}\right)^p\le \left(\int (F+G)^{1/p}\right)^p.$$
Now take $q=1/p$. Then $$\left(\int F^q\right)^{1/q}+\left(\int G^q\right)^{1/q}\le \left(\int (F+G)^q\right)^{1/q}.$$
Hence, what you have is the reverse Minkowski's inequality for $0<q<1$. 
To prove it you first prove that
$$ab\ge \frac{a^q}q-\frac{b^{-r}}{r}$$
for $a\ge 0$ and $b>0$ where $\frac1q-\frac1r=1$. If $a>0$ and $b>0$ then you have equality iff $a^q=b^{-r}$.
Using this you prove the reverse Holder inequality 
$$\int FG\ge \left(\int F^q\right)^{1/q}\left(\int G^{-r}\right)^{-1/r}$$
where $G(x)>0$ for every $x$. Again you have equality iff $(G(x))^{-r}=c(F(x))^q$ for a.e. $x$.
Then you prove Minkoski's inequality in the usual way
 \begin{align}\int (F+G)^q&=\int_{\{F+G>0\}} (F+G)^q=\int_{\{F+G>0\}} (F+G) (F+G)^{q-1}\\&=\int_{\{F+G>0\}} F(F+G)^{q-1}+\int_{\{F+G>0\}} G (F+G)^{q-1}\\
&\ge \left(\int F^q\right)^{1/q}\left(\int (F+G)^{-r(q-1)}\right)^{-1/r}+\left(\int G^q\right)^{1/q}\left(\int (F+G)^{-r(q-1)}\right)^{-1/r}\\&=\left[\left(\int F^q\right)^{1/q}+\left(\int G^q\right)^{1/q}\right]\left(\int (F+G)^{q}\right)^{-1/r}.\end{align}
As in the standard case you equality p>1 you get $F^q=c(F+G)^q$ a.e. and 
$G^q=d(F+G)^q$ a.e. so $F=aG$ a.e. I skipped a lot of details.....
