<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] $b^5=\frac{2-a^5}{a^5},$ where $0<a<\sqrt[5]{2}$ and we need to prove that
$$2a^3+\left(\frac{2-a^5}{a^5}\right)^{\frac{3}{5}}\geq3$$ and since for $3-2a^3\leq0$ the inequality is obvious, we need to prove that $f(a)\geq0$ for any $0<a<\sqrt[3]{\frac{3}{2}},$ where $$f(a)=\frac{3}{5}\ln(2-a^5)-3\ln{a}-\ln(3-2a^3).$$
We see that
$$f'(a)=-\frac{3a^4}{2-a^5}-\frac{3}{a}+\frac{6a^2}{3-2a^3}=$$
$$=\frac{3(a-1)(3+3a+3a^2-a^3-a^4-a^5-a^6-a^7)}{a(2-a^5)(3-2a^3)}$$ and since
$$3+3a+3a^2-a^3-a^4-a^5-a^6-a^7>0$$ for any  $0<a<\sqrt[3]{\frac{3}{2}}$, we obtain $a_{min}=1$ and we are done.
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