Proving a convoluted proof to an inequality: $x,y>0$ such that $x^2+y^2=1$. Prove that, $x^3+y^3 \geqslant \sqrt2 xy $ As the title described, I was trying to find an alternative proof to

If $x,y$ are positive numbers such that $x^2 + y^2=1$, prove that $x^3 + y^3 \geqslant \sqrt2 xy $.

Here's the proof that I've found (I'm sorry, I forgot where I got it):

Apply the Chebyshev's inequality on the tuplets $(x^2, y^2)$ and $\left( \frac1y, \frac1x\right)$, we have $$ \frac12 \left( \frac{x^2}y + \frac{y^2}x \right) \geqslant \frac{x^2 + y^2}2 \cdot \frac{1/y + 1/x}2 \quad \Rightarrow \quad \frac{x^2}y + \frac{y^2}x \geqslant \frac12 \left(\frac1x + \frac1y \right)$$ Apply AM-HM inequality on the tuplets $(x,y)$, we have $$ \frac{x+y}2 \geqslant \frac2{1/x + 1/y} \quad \Rightarrow \quad \frac1x + \frac1y \geqslant \frac4{x+y} $$ Apply Cauchy-Schwartz inequality on the tuplets $(x,y)$ and $(1,1)$, we have $$ x + y = x \cdot 1 + y\cdot 1 \leqslant \sqrt{x^2 + y^2} \sqrt2 = \sqrt2 $$ Combining these 3 inequalities above yield $$ \dfrac{x^2}y + \dfrac{y^2}x \geqslant \frac12 \cdot \frac4{x+y} \geqslant \frac12 \cdot \frac4{\sqrt2} = \sqrt2 $$ The result follows.

Now since I love to punish myself, I tried to find a harder proof as such:

We can let $(x,y) = (\cos\theta, \sin\theta) $, where $\theta \in (0, \tfrac\pi2) $. The inequality in question becomes $$ \begin{array} {l c l }
\cos^3 \theta + \sin^3 \theta &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(\cos^2 \theta - \sin \theta \cos \theta + \sin^2\theta) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(1 - \sin \theta \cos \theta ) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
\cos\theta + \sin\theta &\geqslant & \cos \theta \sin \theta ( \sqrt2 + \cos \theta + \sin \theta) \\
\dfrac1{\sin\theta \cos\theta} &\geqslant & \dfrac{\sqrt2}{\cos \theta \sin \theta} + 1 \\
\end{array}$$ Apply Weierstrass substitution ($t = \tan\frac\theta2$, where $0<t<1$) yields $$ \dfrac{(1+t^2)^2}{2t(1-t^2)} \geqslant \dfrac{\sqrt2 (1+t^2)}{(1-t^2)  +2t} + 1
$$ which simplifies to $$ - \dfrac{t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1}{ 2t(t-1)(t+1) (t^2 - 2t - 1)} \geqslant 0 $$ or $$ t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1 \leqslant 0, \quad\quad\quad 0<t<1$$

Now how do I prove the sextic polynomial inequality above (which is true)?

(This has a fairly obvious improvement
of the inequality.)
Here is a simple algebraic proof of the stronger inequalities
$x^2+y^2=1 
\implies x^3+y^3
\ge \dfrac{x+y}{2} \ge \sqrt{xy}.$
More generally,
without the restriction on
$x^2+y^2$,
this gives
$x^3+y^3
\ge \dfrac{(x+y)(x^2+y^2)}{2}
$.
Proof that this inequality is stronger:
$xy \le (x^2+y^2)/2 = 1/2$
so
$1/\sqrt{xy} \ge \sqrt{2}$.
Therefore
$\sqrt{xy} = xy/\sqrt{xy}
\ge xy \sqrt{2}$.
Proof of the inequality.
$\begin{array}\\
x^3+y^3
&=(x+y)(x^2-xy+y^2)\\
&= (x+y) \dfrac{x^2+y^2+x^2-2xy+y^2}{2}\\
&= (x+y) \dfrac{1+(x-y)^2}{2}\\
&\ge \dfrac{x+y}{2}\\
&\ge \sqrt{xy}\\
\end{array}
$
In all these inequalities, there is equality when x=y and strict inequality otherwise.
Example.
If
$x=3/5, y=4/5$
then
$x^3+y^3 = (27+64)/125 =91/125=0.728,\\
\dfrac{x+y}{2} = 7/10 =0.7\\
\sqrt{xy} = \sqrt{12/25} = 2 \sqrt{3}/5 = 0.6928...,\\
xy \sqrt{2} = 0.6788...
$
