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.
 A: $b^5=\frac{2-a^5}{a^5},$ 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.
A: 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.
A: 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.
