How do I solve this equation: $\sqrt[19]{x^2+y^2-x}+\sqrt[19]{y^2-x^2+x}=2\sqrt[19]{y^2}$?

Question

A very cool equation in my opinion and, in general, it seems to me that I solved it in two ways and seems to be correct:

1. You can consider the function: $f(x)=x^{1/19}$. It grows monotonically and is odd (antisymmetric with respect to zero). If you call: $y^2=a$, $x^2-x=b$. Then your equation will take the form: $f(a+b)+f(a-b)=2f(a)$ Or $f(a+b)-f(a)=-[f(a-b)-f(a)]$. That is, when shifting from point $a$ by distance $b$ to the left and to the right, the function should change by the same amount, but with different signs. It should be obvious if we draw it, then: $1)$ Relatively $0$, the function $f$ is antisymmetric, therefore, for $a=0$, $b$ is any, i.e: $y = 0$ and $x$ - any. $2)$ With respect to any point other than $0$, the function is not antisymmetric, therefore, equality is possible only at zero offset: $a$ - any, $b=0$ or: $y$ - any, $x = 0$ or $x = 1$

2. Due to Jensen's inequality, the left-hand side can be estimated: $$\sqrt[19]{y^2+x^2-x}+\sqrt[19]{y^2-x^2+x}=2 \cdot (0.5\sqrt[9]{y^2+x^2-x}+0.5\sqrt[19]{y^2-x^2+x})≤2 \cdot \sqrt[19]{0.5(y^2+x^2-x+y^2-x^2+x)}=2 \cdot \sqrt[19]{y^2}$$

Moreover, equality is achieved when $$y^2+x^2-x=y^2-x^2+x$$ $$x^2-x=0$$ $x=0$, $x = 1$, $y$ - any.

Answer 1

Hint:

Let $\sqrt[19]{x^2+y^2-x}=a,\sqrt[19]{y^2-x^2+x}=b$

$$a^{19}+b^{19}=2y^2$$

$$a+b=2\left(\dfrac{a^{19}+b^{19}}2\right)^{1/19}\implies\left(\dfrac{a+b}2\right)^{19}=\dfrac{a^{19}+b^{19}}2$$

But we know $$\left(\dfrac{a+b}2\right)^n\le\dfrac{a^n+b^n}2$$

So, we need $a=b\implies x^2+y^2-x=y^2-x^2+x$