Prove that $\left|\frac{x^3y^3}{9x^4+y^4}\right| \le \frac{x^2+y^2}{6}$

Question

I need to prove that: $$\left|\frac{x^3y^3}{9x^4+y^4}\right| \le \frac{x^2+y^2}{6}$$

I am new to inequalities so I only tried C-S and AM-GM but none of those work.

Any hints on how to proceed here?.

Answer 1

Let $y=mx$, then the inequality becomes $$m^4-6m^3+10m^2+9 \ge 0 ~~~~~~(1)$$ Let $f(m)=m^4-6m^3+10m^2+9 \implies f'(m)=4m^3-18m^2+20m=0, m=0,2,5/2$, $f(m)$ has min at $m=0, 5/2$ and max at $m=2$, but $f(0), f(2), f(5/2)>0.$ So $f(m)>0, \forall m \in R$. Equivalently, $f(m)=0$ does not have any real root. Hence the inequality is proved and the equality does not hold.

Answer 2

$(9x^4 + y^4)(x^2 + y^2)\ \ge\ (3x^3 + y^3)^2\ \ge\ |6x^3y^3|$

Answer 3

For $xy=0$ it's obvious.

But for $xy\neq0$ by AM-GM we obtain: $$\left|\frac{x^3y^3}{9x^4+y^4}\right|\leq\frac{|x^3y^3|}{2\sqrt{9x^4y^4}}=\frac{|xy|}{6}\leq\frac{\frac{x^2+y^2}{2}}{6}\leq\frac{x^2+y^2}{6}.$$