Inequatlity for positive functions Let $a \geq 0$, $b \geq 0$, $g \geq 0$. Let $p \geq 1$, $q \geq 1$. Does the following hold:
If $a+b \geq g $ then there exists $C$ independent on $a$ or $b$ such that $a^p+b^q \geq C \min(g^p, g^q)$?
 A: I think we can do it by contradiction. I know that (answer by Surb) $$a^p + b^p \geq C^p (a+b)^p.$$
Now if my claim is not true then for each fixed $g>0$ $\forall i$ exists $a_i$, $b_i$ such that $a_i + b_i \geq g$ and $a_i^p + b_i^q < \frac{1}{i} min(g^p, g^q)$. Putting two inequalities together I obtain
$$\frac{1}{i} > \frac{a_i^p + b_i^q}{(a_i+b_i)^p}$$
and
$$\frac{1}{i} > \frac{a_i^p + b_i^q}{(a_i+b_i)^q}$$
which both must hold.
WLOG $q \geq p$ then if $b_i>1$ we get $b_i^q \geq b_i^p$ and from the first inequality we get $\frac{1}{i} > C^p$ for contradiction. 
If $b_i\leq 1$ then if $a_i \leq 1$ then $a_i^p \geq a_i^q$ and from the second inequality we get contradiction.
If $b_i \leq 1$ and $ a_i > 1$ then we use first inequality to get 
$$\frac{1}{i} > \frac{a_i^p + b_i^q}{(a_i+b_i)^p} \geq \frac{a_i^p}{(2a_i)^p} \geq (\frac{1}{2})^p$$
and we have contradiction for i large enough.
A: Let us denote by $\|\cdot\|_p$ the Minkowski $p$-norm.
Wlog let us suppose that $q\geq p$. Since every norms are equivalent in finite dimensional spaces, there exists $C>0$ such that $\|(x,y)\|_p\geq C\|(x,y)\|_1$ for every $(x,y)\in\Bbb R^2$. Then we get
$$a^p+b^q\geq a^p+b^p = \|(a,b)\|_p^p\geq C^p\|(a,b)\|_1^p=C^p(a+b)^p\geq C^p g^p.$$
