<s>[INST] If $x^2+y^2 <1$ and $4y>x^2$, then $2x^2 + (2y-1)^2 <1$ I'm trying to prove the following two inequalities: 


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*If $x^2+y^2 >1$ and $4y>x^2$, then $2x^2 + (2y-1)^2 > 1$ .

*If $x^2+y^2 <1$ and $4y>x^2$, then $2x^2 + (2y-1)^2 < 1$ .


Both $x$ and $y$ are real numbers. The first inequality can be shown as follows: $$2x^2+(2y-1)^2>2(1-y^2)+(2y-1)^2=2(y-1)^2+1 \geq 1$$ (The condition $4y > x^2$ is superfluous)
For the second inequality, I tried: $2x^2 + (2y-1)^2  =2x^2 + (4y^2 - 4y + 1) <x^2 + 4y^2 + 1  $ (using $-4y < -x^2$). However, this seems not so useful.
It can be proved by plotting graph of ellipses $2x^2 + (2y-1)^2 = r$ for parameter $r$, and calculate the values of $r$ for which graphs are touched.

I seek an another proof of the second inequality. In particular, 'algebraic'(anyway) methods are welcomed.
 [/INST] The second statement is actually false. It can be seen from the graph – it is false in the region that is inside the circle, above the parabola and outside the ellipse. Concretely, $(x,y)=(0.8,0.4)$ violates the second statement.
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