<s>[INST] How do I prove this these inequalities, preferably without calculus? I'm trying to solve an inequality which stems from this problem:
If x, y, z are positive real numbers and $x^5+y^5+z^5=3$, prove $\frac{x^8}{y^3}+\frac{y^8}{z^3}+\frac{z^8}{x^3} \ge 3$
I thought that solving this directly would require an inequality I'm not familiar with, so I decided to set $y=ax$ and $z=bx$ (note $a>0$, $b>0$), reducing the inequality to be proven to:
$\frac{1}{a^3}+\frac{a^8}{b^3}+b^8≥1+a^5+b^5$ for all $a, b > 0$.
I tried proving it by proving the function $f$ with $f(a)=\frac{1}{a^3}+\frac{a^8}{b^3}+b^8≥1+a^5+b^5$ is always positive for $a>0$ and all $b>0$. I graphed it out, and noticed it only attained a value of zero when $a=1$ and $b=1$ (for values of b between 0 and 10). Afterwards, it seemed to only increase. However, the 11th degree polynomials are simply too much for me to handle.
So, my questions would be:
How do I prove the original inequality?
And how do I prove the inequality I have 'reduced' the original one to?
An answer without calculus and without advanced inequality identities would be preferred, but I would prefer an answer with them than no answer at all.
 [/INST] Note that by the rearrangement inequality for triples $(x^3,y^3,z^3)$ and $(x^2,y^2,z^2)$ we can conclude that $$y^3x^2+z^3y^2+x^3z^2 \le x^5+y^5+z^5$$
Now by  Cauchy-Schwarz inequality
$$(\frac{x^8}{y^3}+\frac{y^8}{z^3}+\frac{z^8}{x^3})(y^3x^2+z^3y^2+x^3z^2) \ge (x^5+y^5+z^5)^2 \implies$$ $$\frac{x^8}{y^3}+\frac{y^8}{z^3}+\frac{z^8}{x^3} \ge \frac{(x^5+y^5+z^5)^2}{y^3x^2+z^3y^2+x^3z^2} \ge \frac{(x^5+y^5+z^5)^2}{x^5+y^5+z^5} =x^5+y^5+z^5 = 3$$
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