3

I've been trying to prove an inequality I was given by a friend, but so far my only progress has been calculus bashing: $$LHS \ge RHS\iff1+x^3+y^3 - x-x^2y-y^2 \ge0$$ Letting $f(x,y) = 1+x^3+y^3 - x-x^2y-y^2$, we want $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0$ for the minimum.

Hence $$3x^2-1-2xy=0 \ \ \ \ \ \ \ \ ... (1)$$ and $$3y^2-x^2-2y=0 \ \ \ \ \ \ \ \ ... (2)$$ which gives one solution$,(x,y)=(1,1),$ in the first quadrant:

graph

The trouble is that proving that $(x,y)=(1,1)$ is the only positive solution to the system of equations is quite cumbersome. $$\ \ $$

Is there any way we can prove this without calculus, maybe using AM-GM or Cauchy-Schwarz inequalities? I used AM-GM to obtain $$1+x^3+y^3 \ge 3\sqrt[3]{1^3x^3y^3}=3xy,$$ but I'm not sure how this helps me proceed. Many thanks in advance.

Ѕᴀᴀᴅ
  • 34,263
user1050148
  • 218

2 Answers2

6

Note that $$ \begin{align} \frac{1}{3}(1 + 1+x^3) &\ge x \\ \frac{1}{3}(x^3 + x^3 + y^3) &\ge x^2y \\ \frac{1}{3}(1 + y^3 + y^3) &\ge y^2 \end{align} $$

Chia
  • 1,023
  • Thanks for this, that is a clever method. I can see how these equations add to produce the original inequality. – user1050148 Dec 21 '22 at 09:18
4

Let $1=z^3$.

Thus, we need to prove that: $$x^3+y^3+z^3\geq x^2y+y^2z+z^2x,$$ which is true by Rearrangement.

Also, we have: $$\sum_{cyc}(x^3-x^2y)=\sum_{cyc}\left(x^2(x-y)-\frac{1}{3}(x^3-y^3)\right)=\frac{1}{3}\sum_{cyc}(x-y)^2(2x+y)\geq0.$$

Michael Rozenberg
  • 194,933