<s>[INST] Conditional inequality $2a^3+b^3≥3$ 
Non-negative $a$ and $b$ such that $a^5+a^5b^5=2$. How then do I prove
the following inequality $2a^3+b^3≥3$?

So, we can try using the Lagrange multiplier method:
Let $f(a, b)=2 a^{3}+b^{3}+\lambda(a^{5}+a^{5} b^{5}-2), \quad a, b \geq 0$
$$\tag1 \frac{\partial f}{\partial a}=6 a^{2}+\lambda(5 a^{4}+5 a^{4} b^{5})=0 \ldots$$
$$\tag2 \frac{\partial f}{\partial b}=3 b^{2}+\lambda(5 a^{5} b^{4})=0 \ldots$$
$$\left\{\begin{array}{c} 6+\lambda(5 a^{2}+5 a^{2} b^{5})=0 \\ 3+\lambda(5 a^{3} b^{4})=0 \end{array}\right. $$
$$ \lambda=-\frac{6}{5 a^{2}+5 a^{2} b^{5}}=-\frac{3}{5 a^{3} b^{4}} $$
Multiply by $a^3$, $\,2 a^{6} b^{4}=a^{5}+a^{5} b^{5}=2$.
$$ a^{6} b^{4}=1,\, a^{3} b^{2}=1 \Rightarrow b=\frac{1}{a^{\frac{3}{2}}}\quad a, b \geq 0 $$
$$ \begin{gathered} a^{5}+a^{5} b^{5}=2 \Rightarrow a^{5}+\frac{a^{5}}{a^{\frac{15}{2}}}=2 \Rightarrow a^{5}+\frac{1}{a^{\frac{5}{2}}}=2 \Rightarrow a^{5}-2 a^{\frac{5}{2}}+1=0 \\ \Rightarrow\left(a^{\frac{5}{2}}-1\right)^{2}=0 \Rightarrow a=1 \end{gathered} $$
And the minimum is at $(a,b)=(1,1)$.
I'm not sure I solved this inequality correctly, I would like to see a more beautiful way.
 [/INST] Remarks: Without calculus.
We have $a^5 = \frac{2}{1 + b^5}$.
We need to prove that $2a^3 \ge 3 - b^3$.
WLOG, assume that $3 - b^3 \ge 0$.
It suffices to prove that
$$2^5 a^{15} \ge (3 - b^3)^5$$
or
$$2^5 \left(\frac{2}{1 + b^5}\right)^3 \ge (3 - b^3)^5$$
or
$$2^9 \ge (1 + b^5)^3 \cdot (3 - b^3)^3 \cdot 2(3 - b^3)^2.$$
By AM-GM, we have
$$b^5 \le \frac{b^3 + b^6 + b^6}{3}$$
and
$$(3 - b^3)\cdot (3 - b^3)\cdot 2
\le \left(\frac{3 - b^3 + 3 - b^3 + 2}{3}\right)^3.$$
Thus, it suffices to prove that
$$2^9 \ge \left(1 + \frac{b^3 + b^6 + b^6}{3}\right)^3\cdot (3 - b^3)^3 \cdot \left(\frac{3 - b^3 + 3 - b^3 + 2}{3}\right)^3$$
or
$$2^3 \ge \left(1 + \frac{b^3 + b^6 + b^6}{3}\right)\cdot (3 - b^3) \cdot \left(\frac{3 - b^3 + 3 - b^3 + 2}{3}\right)$$
or
$$\frac{2}{9}b^3(9 - 2b^3)(b^3 - 1)^2 \ge 0$$
which is true.
We are done.
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