How to prove this: $a^4+b^4+2 \ge 4ab$? How to prove this: $a^4+b^4+2 \ge 4ab$?
$a$ and $b$ are reals.
 A: There's a fast way to apply the arithmetic-geometric mean inequality:
We split up $2$ into $1+1$ and get
$$\frac{a^4+b^4+1+1}{4}\ge\sqrt[4]{a^4b^4}=ab$$
Rearranging, we are done.
A: Notice that:
$$(a^2+b^2)^2 = a^4 + b^4 + 2a^2b^2$$
So, then we have that: $(a^2 + b^2)^2 - 2a^2b^2 + 2 \ge 4ab$
It follows that:
$$(a^2 + b^2)^2 \ge 2a^2b^2 + 4ab +2 -4,\ (a^2 + b^2)^2 \ge 2(ab+1)^2 -2^2 = \left(\sqrt{2}(ab+1) + 2\right)\left(\sqrt{2}(ab+1) - 2\right)$$
Our inequality now becomes:
$$(a^2 + b^2)^2 \ge\left(\sqrt{2}ab+\sqrt{2} + 2\right)\left(\sqrt{2}ab+\sqrt{2} - 2\right)$$
NOTE This manipulation may or may not assist you in proving the inequality. I only provided it as a possible way to help you in your proof.
A: Another possible method is generation of a function and calculating its minimum:
$$f(a,b)=a^4+b^4+2-4 a b$$ 
$$D_a f(a,b)=4 a^3-4 b = 0$$ 
$$ a^3=b$$ 
$$ f(a)=a^4+a^{12}+2-4a^4=a^{12}+2-3a^4$$ 
$$D_a f(a)=12a^{11}-12a^3=0$$ 
$$a^3 (a^8-1)=0$$ 
Minimum must be at one of:
$$f(0,0)=2$$
$$f(1,1)=0$$
$$f(-1,-1)=0$$
so
$$f(a,b)=a^4+b^4+2-4 a b\geq0$$ 
A: Inequalities like this can most often be proved or simplified by using the Arithmetic-Geometric mean inequality.
First apply AM-GM in $a^4$ and $b^4$:
$$\frac{a^4+b^4}{2} \ge \sqrt{a^4b^4}$$
$$a^4+b^4 \ge 2a^2b^2$$
Now apply AM-GM in $2a^2b^2$ and $2$:
$$\frac{2a^2b^2+2}{2} \ge \sqrt{4a^2b^2}$$ 
$$2a^2b^2+2 \ge 4ab$$
I suppose the result is quite obvious now. This answer was, of course, posted with the assumption that $a$ and $b$ are reals.
