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)?
Best Answer
(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... $