Prove that if $0\leq a,b$ and $a+b=1$ then $x^ay^b\leq ax+by$ for $x, y >0$ Would like help getting started, unfamiliar with proving inequalities in general. 
 A: this is a simple and pretty form of a very widely-used result, so it is worth looking at it from many different points of view. here is one -
suppose $b  \in (0,1]$. let $s$ denote a real variable on $[1,\infty)$. and consider the function:
$$
f(s) = 1 + b(s-1) - s^b
$$
clearly $f(1)=0$. and we have
$$
f'(s) = b(1-s^{b-1}) \ge 0
$$
with the inequality strict if $s \gt 1$. 
this means that $f \ge 0 $, which we may write as:
$$
s^b \le 1- b + bs \tag{1}
$$
suppose $y \ge x \gt 0$ and set $s=\frac{y}{x}$. then (1) becomes:
$$
\left(\frac{y}{x} \right)^b \le (1-b) + b\left(\frac{y}{x} \right)
$$
multiplying both sides by $x$ now gives:
$$
x^{1-b}y^b \le (1-b)x + by
$$
writing $a$ for $1-b$ gives the result required
A: Consider, $f(x)=\log(x)$. Then $f''(x)=-\frac{1}{x^2}<0$. So, $f$ is concave function. Since $a+b=1$ so, using Jensen's inequality ,
$$f(ax+by)\ge af(x)+bf(y).$$
$$\implies \log(ax+by)\ge a\log x+b\log y$$
$$\implies ax+by\ge x^ay^b.$$
A: HINT: Concavity of logarithms.
A: Using Jensen's inequality:
$f(\frac{\sum_i a_i x_i}{\sum_i a_i})  \geq \frac{\sum_i a_i f(x_i)}{\sum_i a_i}$ for a concave function $f$.
$f=log(x)$ is concave.
Thus,
$log(\frac{\sum_i a_i x_i}{\sum_i a_i})  \geq \frac{\sum_i a_i log(x_i)}{\sum_i a_i}$ $\implies$ $log(\frac{\sum_i a_i x_i}{\sum_i a_i})  \geq log((\Pi x_i^{a_i})^{\frac{1}{\sum_i a_i}})$ $\implies$ $\frac{\sum_i a_i x_i}{\sum_i a_i}  \geq (\Pi x_i^{a_i})^{\frac{1}{\sum_i a_i}}$
The last inequality is also the genereal weighted AM-GM inequality.
A: It's just AM-GM:
$$ax+by\geq x^ay^b.$$
Done!
