using AM-GM inequality to prove other inequalities Prove that $$a^4 + b^4 + 8 \ge 8ab$$
I've been working on this for an entire day and nothing. I tried using AM-GM inequality. Is there any other inequality theorem that I should be using instead? I need a full proof and explanation. Thanks
 A: Using AM-GM we get
$$
a^4+b^4 + 8 \ge 2(ab)^2 + 8
$$
We want to show that
$$
2(ab)^2 + 8 \ge 8ab
$$
It's equivalent to
$$
2(ab-2)^2\ge 0
$$
which is clearly true
A: You need two applications of AM-GM
$$
\color{blue}{a^4+b^4}+8\ge\color{green}{2a^2b^2}+8\ge2[\color{blue}{(ab)^2+2^2}]\ge2\times\color{green}{2ab\times2}=8ab
$$
A: Recall young's inequality for (three) products:
$$abc\leq\frac{a^p}{p}+\frac{b^q}{q}+\frac{c^r}{r}$$
when $1/p+1/q+1/r=1$. If you haven't seen this, you can easily prove it using the concavity of the logarithm. Your result is immediate when you let $c=2$, $p=q=4$, and $r=2$.
Note that in a particular way, young's inequality is a more general version of am-gm. In this case it is more direct.
A: Just to start from the beginning, the AM-GM inequality is this:
$$\frac{m+n}{2} \geq \sqrt{mn}.$$
We can replace $(m,n)$ with $(a^4, b^4)$:
$$\frac{a^4+b^4}{2} \geq \sqrt{a^4 b^4},$$
or
$$a^4 + b^4 \geq 2a^2b^2.$$
Then adding $8$ to both sides doesn't change anything:
$$a^4 + b^4 + 8 \geq 2a^2b^2 + 8.$$
This gives an inequality that follows from AM-GM that has the left-hand side as what you need. Then you continue as in one of the other answers to show what you need.
