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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 $x^4+y^4+z^4+16$.

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    $\begingroup$ Hint: start by dividing both sides of the inequality by $4$. $\endgroup$ – Daniel Fischer Jul 26 '15 at 20:12
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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.

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  • $\begingroup$ ah, thank you. That was a very sneaky question $\endgroup$ – Andrew Brick Jul 26 '15 at 20:32
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I don't think you are going to make the LHS of your true inquality look like the inquality you are trying to prove.

But since the 8xyz is identical, and in both cases "less than or equal to" the left side.

If you can prove that x^4+y^4+z^4+16>=((x+y+z+2)^4)/3.

then it will stand that since ((x+y+z+2)^4)/3>=8xyz, so is x^4+y^4+z^4+16, which is in fact even greater.

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  • $\begingroup$ Well, supinf's answer is simpler. $\endgroup$ – Carl Jul 26 '15 at 23:07

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