Prove the following inequality from jensen's inequality

By using the concave function $f(x)=\ln(x)$ inside the jensen inequality, I get the result:

$$\sqrt[n]{t_1t_2\cdots t_n}\leq \frac{t_1+\cdots+t_n}{n}$$
Where $t_1,\ldots,t_n\in \mathbb{R}_{>0}$

From this result, I am trying to prove that
$x^4+y^4+z^4+16\geq 8xyz$

My attempt at proving this is as follows, let $n=4$, $t_1=x,t_2=y,t_3=z$ and $t_4=2$, hence:

$$\sqrt[4]{2xyz}\leq\frac{x+y+z+2}{4}$$

$$2xyz\leq\frac{(x+y+z+2)^4}{4^4}$$

$$8xyz\leq\frac{(x+y+z+2)^4}{4^3}$$

But now I have trouble trying to get the upper limit to $x^4+y^4+z^4+16$.

Answer

You almost got it, the solution is to set
$t_1=x^4 , t_2=y^4, t_3 = z^4, t_4=16$.
This gives you

$$2xyz \leq \frac{x^4+y^4+z^4+16}4$$ , which is what you want.