Name and proof of mathematical inequality I'm reading a book on functional analysis and I've come across an inequality i.e.
Suppose we have $$ a, b, c, p, q \ge 0 $$ and $$ p + q = 1 $$ then
$$ a+b^pc^q \le (a+b)^p(a+c)^q$$
I'm curious on how one would prove this inequality let alone if there is a name of this inequality.  I'd like to note this is not a homework problem.  Thanks!
 A: If $a=0$ or $b=0$ or $c=0$ then it is obvious. Let $a,b,c\neq 0$. Divide by $b^pc^q$ and get:
$$\frac{a^pa^q}{b^pc^q}+1\leq (\frac{a}{b}+1)^p(\frac{a}{c}+1)^q$$
Now substitute $x=\frac{a}{b}> 0$ and $y=\frac{a}{c}>0 $. We have to prove 
$$x^py^q+1\leq (x+1)^p(y+1)^q$$
Note that $\frac{1}{\frac{1}{p}}+\frac{1}{\frac{1}{q}}=1$ and use Holder's inequality:
$$x^py^q+1\cdot1\leq ((x^p)^{1/p}+1^{1/p})^p((y^q)^{1/q}+1^{1/q})^q$$ 
Note: Direct application of Holder's inequality also yields the result
$$a+b^pc^q=a^p\cdot a^q+b^p\cdot c^q\leq ((a^p)^{1/p}+(b^p)^{1/p})^p((a^q)^{1/q}+(c^q)^{1/q})^q=(a+b)^p(a+c)^q$$
A: If $a=0$, the inequality is trivial. Suppose now $a>0$. Then we can divide the inequality by $a$, reducing it to considering
$$ 1+x^p y^q \leqslant (1+x)^p(1+y)^q \tag{1} $$
for $x,y \geqslant 0$. If one of $x,y$ is zero, then the inequality reduces to one of the form
$$ 1 \leqslant (1+x)^p, $$
again trivial. Likewise for $p=1,q=0$. At this point it's probably sensible to divide everything by the right-hand side of (1), to reduce to 
$$ \alpha^p \beta^q + (1-\alpha)^p (1-\beta)^q \leqslant 1, $$
where $\alpha=(1+x)^{-1}$ and $\beta=(1+y)^{-1}$, so $0<\alpha,\beta \leqslant 1$.
Now we can use the AM–GM inequality: recall that if $\sum_i a_i=1$,
$$ \sum_i a_i x_i \geqslant \prod_i x_i^{a_i} $$
In this case we have
$$ \sum_i x_i^p y_i^q \leqslant \sum_i (px_i + qy_i) = 1, $$
since $\sum_i x_i = \sum_i y_i = 1$, which gives the required inequality when we stick in $x_1=\alpha=1-x_2$, $y_1=\beta=1-y_2$.
