Prove the following inequality for any positive real number $x,y$

Question

Prove the following inequality for any positive real number $x,y$

$xy^3 \leq \frac14x^4 + \frac34y^4$

I have tried subtracting $y^4$ from both sides to get:

$xy^3 - y^4 \leq \frac14x^4 - \frac14y^4$

$y^3(x - y) \leq \frac14(x - y)(x + y)(x^2 + y^2)$

$y^3 \leq \frac14(x + y)(x^2 + y^2)$

$0 \leq (x + y)(x^2 + y^2) - 4y^3$

$0 \leq x^3 + x^2y +xy^2 - 3y^3$

But I don't know where to go from here since I couldn't find a way to conveniently factor this. I have tried completing the square from this point, but I couldn't see any obvious way to prove the right side to be greater than zero.

I also tried completing the square from the beginning to get:

$0 \leq x^4 + 3y^4 - 4xy^3$

$0 \leq x^4 + 3y^2(y - \frac23x)^2 -\frac43x^2y^2$

But yet again I couldn't figure out where to go from here. Let me know what I'm doing wrong or what other methods I can try. Thanks!

Answer 1

Using AM-GM: $$\frac14x^4 + \frac34y^4=\frac14x^4+\frac14y^4+\frac14y^4+\frac14y^4\ge 4\sqrt[4]{\frac1{4^4}x^4y^{12}}=xy^3.$$

Answer 2

Using algebra, let $y=kx$ and since $x \neq 0$, the problem is to show that $$k^3 < \frac 14+\frac 34 k^4$$ that is to say that the function $$f(k)=3k^4-4k^3+1$$ is always positive.

We have $$f'(k)=12k^3-12k^2=12k^2(k-1)$$ The derivative only cancels for $k=1$ since, as you wrote, $x$ and $y$ are not zero. Now, $f(1)=0$.

The second derivative being $$f''(k)=36k^2-24k$$ gives $f''(1)=12$ then, by the second derivative test, $0$ is the minimum value of $f(k)$ and the inequality holds.

Answer 3

We need to prove that $$x^4-4xy^3+3y^4\geq0$$ or $$x^4-2x^3y+x^2y^2+2x^3y-4x^2y^2+2xy^3+3x^2y^2-6xy^3+3y^4\geq0$$ or $$(x-y)^2(x^2+2xy+3y^2)\geq0,$$ which is true even for all reals $x$ and $y$.

Answer 4

Another way of proving this is to apply Young's inequality for $p=4$ and $q=\frac{4}{3}$ which implies $$ xy^3\leq \frac{x^p}{p}+\frac{(y^3)^q}{q}=\frac{1}{4}x^4+\frac{3}{4}y^4. $$