I have these two problems among other similar question that I haven't figured out how to solve. I believe I could use the AM–GM inequality to solve this but I'm unsure how that would look.
Let $a$ and $b$ be two non-negative real number and let $n$ be a positive integer. Show that
$$(a+b)^n\leq 2^{n-1}(a^n+b^n)$$
Let $a$, $b$ and $c$ be three positive real numbers. Show that
$$\frac{a^3+b^3+c^3}{a^2+b^2+c^2}\geq \frac{a+b+c}{3}$$
Thanks
AM-GM works for the first inequality if you apply it as the Power Mean Inequality, that is:
$$\sqrt[n]{\frac{a^n+b^n}{2}}\geq \frac{a+b}{2}$$
and raise to the $n$th power. You can also notice that it is direct Holder:
$$(a^n+b^n)(1^n+1^n)\cdot ... \cdot(1^n+1^n) \geq (a+b)^n$$
The second inequality is also a direct application of the Chebyshev Inequality because the sequences $(a,b,c)$ and $(a^2,b^2,c^2)$ are similary ordered, so
$$\frac{1}{3}(a+b+c)\cdot \frac{1}{3}(a^2+b^2+c^2) \leq \frac{1}{3}(a\cdot a^2+b\cdot b^2+c\cdot c^2)$$
implying the desired result. Another idea is to expand the inequality as:
$$2(a^3+b^3+c^3) \geq ab(a+b)+bc(b+c)+ca(c+a)$$
and this can be written as:
$$(a+b)(a-b)^2+(b+c)(b-c)^2+(c+a)(c-a)^2 \geq 0$$
or proven via AM-GM:
$$a^3+a^3+b^3 \geq 3a^2b$$
$$a^3+b^3+b^3 \geq 3ab^2$$
Now sum to get
$$a^3+b^3\geq ab(a+b)$$
and sum cyclically to arrive at the expanded inequality