Is it true that $a^p b^q \leq a+b$ with $p + q = 1$? Let $a,b \geq 0$ and $0<p,q < 1$ s.t. $p + q = 1$.
Is it true that $a^p b^q \leq a+b$?
 A: Regarding the edited question, assume w.l.o.g. That $a \leq b$.
Then
$$a^p b^q \leq b^p b^q = b \leq a+b.$$
A: The comments given are counter examples. A famous inequality that resembles yours is
$a,b,p,q$ are positive real numbers and $\frac{1}p+\frac{1}q = 1$ then,
$$\frac{a^{p}}{p} + \frac{b^{q}}{q} \ge ab.$$
A: Well, the inequality is much stronger. In fact we have $a^{p}b^{q} \leq pa + qb$ and equality holds only when $a = b$. Since $0 < p, q < 1$ it follows that $pa < a, qb < b$ so that $pa + qb < a + b$ and therefore we get $a^{p}b^{q} \leq pa + qb < a + b$. The proof for the general inequality $a^{p}b^{q} \leq pa + qb$ can be given by writing $q = 1 - p$. Then we need to prove for $a, b \geq 0$ and $0 < p < 1$ that $$a^{p}b^{1 - p} \leq ap + b(1 - p)$$ clearly we can assume $b > a$ and consider the function $f(x) = x^{1 - p}$ so that $f'(x) = (1 - p)x^{-p}$. Then by mean value theorem we have $$f(b) - f(a) = (b - a)f'(c)$$ for some $c \in (a, b)$. This means that $$b^{1 - p} - a^{1 - p} = (1 - p)(b - a)c^{-p}$$ We have $a < c < b$ therefore $c^{-p} < a^{-p}$ (note that $(-p) < 0$). And thus we get $$b^{1 - p} - a^{1 - p} < (1 - p)(b - a)a^{-p}$$ and multiplying by $a^{p} > 0$ we get $$a^{p}b^{1 - p} - a < (1 - p)(b - a)$$ or $$a^{p}b^{1 - p} < pa + (1 - p)b$$
A: Since  for $1/q+1/q=1$
$$
\frac{a^p}{p}+\frac{b^q}{q}\ge ab
$$
We can redefine $\tilde p=1/p$, $\tilde q=1/q$, $\tilde a=a^p$, $\tilde b =b^q$
Then the above inequality implies 
$$
\tilde a+\tilde b\ge p \tilde a +q \tilde b \ge \tilde a ^{\tilde p} b^{\tilde q}
$$
REMARK:  I believe $p,q>0$ and not $>1$ as written. 
