<s>[INST] Algebraic inequalities on different means If $a^2+b^2+c^2+d^2=1$ in which $a,b,c,d>0$, prove or disprove
\begin{equation*}
\begin{aligned}
(a+b+c+d)^8&\geq 2^{12}abcd;\\
a+b+c+d+\frac{1}{2(abcd)^{1/4}}&\geq 3.
\end{aligned}
\end{equation*}
Can you tell any general algorithm for this type of problems? Thanks.
 [/INST] There is a general algorithm for this type of problems.
We can use the Vasc's EV Method: https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf
For example, your second inequality is wrong, but the following is true.

Let $a$, $b$, $c$ and $d$ be positive numbers such that $a^2+b^2+c^2+d^2=1$. Prove that:
  $$a+b+c+d+\frac{2}{3\sqrt[4]{abcd}}\geq\frac{10}{3}.$$

Indeed, let $a+b+c+d=constant.$
Thus, it's enough to prove our inequality for a maximal value of $abcd$, which  by Corollary 1.8(b) 
happens for equality case of three variables.
After homogenization we need to prove that
$$a+b+c+d+\frac{2(a^2+b^2+c^2+d^2)}{3\sqrt[4]{abcd}}\geq\frac{10}{3}\sqrt{a^2+b^2+c^2+d^2}.$$
Now, let $a=x^4$, where $x>0$ and $b=c=d=1$.
Thus, we need to prove that
$$x^4+3+\frac{2(x^8+3)}{3x}\geq\frac{10}{3}\sqrt{x^8+3}$$ or after squaring of the both sides
$$(x-1)^2(4x^{14}+8x^{13}+12x^{12}+28x^{11}+44x^{10}+60x^9-15x^8-54x^7-69x^6-84x^5-45x^4+30x^3+105x^2+180x+36)\geq0,$$ which is obvious by AM-GM.
By this method you can create infinitely many inequalities. 
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