How to prove $x^4+2x^2y^2+y^4\geq2xy^3$

Question

Suppose that $x,y$ are real numbers. I want to prove $$x^4+2x^2y^2+y^4\geq2xy^3.$$ I noticed that this is the same as $$(x^2+y^2)^2\geq 2xy^3.$$ Can we proceed from here?

Answer 1

We know that: $$(x^2+y^2)\ge 2xy$$

Multiplying both sides by $y^2$

$$(x^2+y^2)\cdot y^2\ge 2xy^3$$

And clearly $x^2+y^2 \ge y^2$. So

$$(x^2+y^2)\cdot(x^2+y^2)\ge(x^2+y^2)\cdot y^2\ge 2xy^3$$

implying that $$\boxed{(x^2+y^2)^2 \ge 2xy^3}$$

Answer 2

Hint: $x^2 + y ^2 \ge 2xy, x^2 + y^ 2 \ge y^ 2$ .

Answer 3

We can assume without loss of generality that $x,y\ge 0$ (why?)

So by AM-GM inequality:

$$x^2+y^2=x^2+\frac{y^2}3+ \frac{y^2}3+ \frac{y^2}3\ge4\sqrt[4]{\frac{1}{3^3}x^2y^6}=4\cdot3^{-\frac34}\sqrt{xy^3}.$$

By squaring:

$$(x^2+y^2)^2\geq16\cdot3^{-\frac32}xy^3\approx3.08 xy^3.$$

So we have the sharper bound $$\boxed{(x^2+y^2)^2\geq 3xy^3.}$$

Answer 4

Consider :

$r>0$, $0\le \theta \lt 2π$.

Then obviously:

$r^4 \ge r\sin (2\theta) r^3 \sin^2 (\theta)$(Why?)

$r^4 \ge 2r \sin (\theta) \cos (\theta) r^3 \sin^2 (\theta)$

And with $x=r\cos (\theta)$, $y=r \sin (\theta)$:

$(x^2+y^2)^2\ge 2 xy^3.$

Answer 5

I think the following is a bit of better than your attempt. $$x^4+2x^2y^2-2xy^3+y^4=x^4+x^2y^2+y^2(x-y)^2\geq0.$$