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.

Proof by contradiction. Suppose $2a^3 + b^3 < 3$.
Then $ 0 \leq a \leq \sqrt[3]{3/2}$ and $a^5 + a^5b^5 < a^5 + a^5 (3-2a^3)^{5/3} $.
Let $ f(a) = a^5 ( 1 + (3-2a^3)^{5/3})$.
Verify that $f'(a) = 5a^4 ( 1 + (3-4a^3) (3-2a^3)^{2/3})  $.
Verify that $f'(a)$ is $0$ on $\{ 0, 1 \}$, positive on $(0, 1)$, and  negative on $(1, \sqrt[3]{3/2}]$.
So $f(a)$ achieves its maximum at $a = 1$.
Thus, $f(a) \leq 2$.
So $a^5 + a^5b^5 < f(a) \leq 2 $, which is a contradiction.

I wanted to do this without calculus, but wasn't successful.
