Prove that for any non-zero $x, y$, one has $x^6/y^2+y^6/x^2>x^4+y^4$.

Question

I removed the denominators, so the equation became $x^8+y^8 \geq x^6y^2+y^6x^2$. Obviously, if $x=y$, then the two equations would be equal to each other, but I have trouble proving what would happen when $x$ didn't equal $y$.

Answer 1

Suppose wlog $$x^2>y^2 \implies \frac1{x^2}<\frac1{y^2}$$

thus by Rearrangement inequality

$$\frac{x^6}{y^2}+\frac{y^6}{x^2}>\frac{x^6}{x^2}+\frac{y^6}{y^2}=x^4+y^4$$

Answer 2

\begin{eqnarray*} ( (2x^2+y^2)^2 +3y^4)(x^2-y^2)^2 \geq 0. \end{eqnarray*}

Answer 3

We can assume $x,y > 0$.

Since the inequality is homogeneous, we can scale the variables, hence we can assume $y=1$.

Then, letting $t=\sqrt{x}$, the inequality reduces to $$t^4 -t^3 -t + 1 \ge 0$$ which holds true since the LHS factors as $$(t^2+t+1)(t-1)^2$$

Answer 4

Dividing by $\,x^2y^2 \gt 0\,$ gives the equivalent inequality:

$$\frac{x^4}{y^4}+\frac{y^4}{x^4} \ge \frac{x^2}{y^2}+\frac{y^2}{x^2} \;\;\iff\;\; 0 \le t^2 - t - 2 = (t+1)(t-2) \quad\text{with}\;\; t = \frac{x^2}{y^2} + \frac{y^2}{x^2} \ge 2$$

Answer 5

WLOG we can assume $x,y>0$ $$x^8+2x^4y^4+y^8\geq x^6y^2+2x^4y^4+x^2y^6\\(x^4+y^4)^2\geq x^2y^2(x^4+2x^2y^2+y^4)\\(x^4+y^4)^2\geq x^2y^2(x^2+y^2)^2\\x^4+y^4\geq xy(x^2+y^2)\\x^4-x^3y-y^3x+y^4\geq 0\\x^3(x-y)-y^3(x-y)\geq 0\\(x-y)(x^3-y^3)\geq 0\\(x-y)^2(x^2+xy+y^2)\geq 0$$ Which is always true since $x^2+xy+y^2\geq 0$