I'm trying to prove that the sum of two log-convex functions is log-convex. I've figured out that this can be done by proving:
$a^\alpha b^\beta + c^\alpha d^\beta \leq (a+c)^\alpha (b+d)^\beta$
for $a,b,c,d,\alpha,\beta \in \mathbb{R}$ and $\alpha,\beta \geq 0$ (actually, in my case $\alpha + \beta = 1$, in addition). Any ideas?
Edit: Assume further that $a,b,c,d > 0$ (in the original problem, they are inputs to a logarithm and so must be positive).
Hint:Assume further that $a,b,c,d > 0$. Put $x = \dfrac{a}{a+c}, y = \dfrac{b}{b+d}$. The inequality is now:
$x^{\alpha}y^{\beta} + (1-x)^{\alpha}(1-y)^{\beta} \leq 1$. Consider $LHS = f(x,y)$. Can you use this to continue? by finding critical values of $f$: $f_x = f_y = 0$?
Hint:
Divide both sides by $a^{\alpha}b^{\beta}$ and let $x=c/a$ and $y=d/b$. The inequality is equivalent to
$$1\le (1+x)^{\alpha}(1+y)^{\beta}-x^{\alpha}y^{\beta}$$
WLOG, due to homogeneity we may set $a+c=b+d=1$. Now we need to show $$ a^\alpha b^\beta +c^\alpha d^\beta \le 1$$
which follows from summing the AM-GMs $$a^\alpha b^\beta \le \alpha a + \beta b, \qquad c^\alpha d^\beta \le \alpha c + \beta d$$
