Let $a$ and $b$ be two positive numbers such that $a+b=1$. I am supposed to show that $u^av^b\leq au+bv$ for all positive $u$ and $v$.
It is known that $\ln x \leq x$ for all positive $x$, so I managed to get $\ln(u^av^b)=a\ln(u)+b\ln(v)\leq au+bv$ instead. Not sure if this direction is worth exploring or did I miss the point altogether?
First, let's consider the case if $a,b\in \mathbb Q$.n Let $a=\frac{m}{m+n},b=\frac{n}{m+n}$, $m,n\in \mathbb N$.
Then we only need to prove that $mu+nv\geq(m+n)u^{\frac{m}{m+n}}\cdot v^{\frac{n}{m+n}}$
By AM-GM, $mu+nv=u+...+u+v+...+v\geq(m+n)(u\cdot u \cdot\cdot\cdot u\cdot v\cdot\cdot\cdot v)^{\frac{1}{m+n}}=(m+n)u^{\frac{m}{m+n}}\cdot v^{\frac{n}{m+n}}$. Done!
(In the first equation there are $m+n$ terms, so use AM-GM like this:$$x_1+...+x_{m+n}\geq(m+n)(x_1\cdots x_{m+n})^{\frac{1}{m+n}}$$)
Because of the continuity of $a,b$, so this inequality holds for $a,b\in \mathbb R$.
In fact, this inequality is the weighted form of AM-GM.
