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Did
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We may see that the inequality is true for every $|x|<0.597$ in the following way:

For a given value of $x$ consider the values of $N$ and $n$. The inequality will be true for this $x$ if the quadratic polynomial in $y$ $$(y^2+1)N+y\, n-(y N+n)^2-N^2 >0$$$$(y^2+1)N+y\, n-(y N+n)^2-N^2$$ is always positive. In other words the inequality is true for this $x$ isas soon as the discriminant $\Delta$ of this quadratic is $<0$.negative (Thethe coefficient of y^2 is$y^2$ being positive).

If I am not wrong theThe discriminant is equal to $\Delta =n^2-4N^2(1-N)^2$. Since $n^2<\frac{1}{2\pi}$$n^2<1/(2\pi)$, the inequality will be true for every $x$ such that $4N^2(1-N)^2>1/2\pi$$4N^2(1-N)^2>1/(2\pi)$.

This gives, thatThus the inequality is true whenfor every $x$ such that $0.275214<N<0.724786$. This correspondcorresponds to the condition $|x|<0.597$.

We may see that the inequality is true for $|x|<0.597$ in the following way:

For a given value of $x$ consider the values of $N$ and $n$. The inequality will be true for this $x$ if the quadratic polynomial in $y$ $$(y^2+1)N+y\, n-(y N+n)^2-N^2 >0$$ is always positive. In other words the inequality is true for this $x$ is the discriminant of this quadratic is $<0$. (The coefficient of y^2 is positive).

If I am not wrong the discriminant is equal to $\Delta =n^2-4N^2(1-N)^2$. Since $n^2<\frac{1}{2\pi}$, the inequality will be true for $4N^2(1-N)^2>1/2\pi$.

This gives, that the inequality is true when $0.275214<N<0.724786$. This correspond to $|x|<0.597$.

We may see that the inequality is true for every $|x|<0.597$ in the following way:

For a given value of $x$ consider the values of $N$ and $n$. The inequality will be true for this $x$ if the quadratic polynomial in $y$ $$(y^2+1)N+y\, n-(y N+n)^2-N^2$$ is always positive. In other words the inequality is true for this $x$ as soon as the discriminant $\Delta$ of this quadratic is negative (the coefficient of $y^2$ being positive).

The discriminant is $\Delta =n^2-4N^2(1-N)^2$. Since $n^2<1/(2\pi)$, the inequality will be true for every $x$ such that $4N^2(1-N)^2>1/(2\pi)$.

Thus the inequality is true for every $x$ such that $0.275214<N<0.724786$. This corresponds to the condition $|x|<0.597$.

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juan
juan
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We may see that the inequality is true for $|x|>0.597$$|x|<0.597$ in the following way:

For a given value of $x$ consider the values of $N$ and $n$. The inequality will be true for this $x$ if the quadratic polynomial in $y$ $$(y^2+1)N+y\, n-(y N+n)^2-N^2 >0$$ is always positive. In other words the inequality is true for this $x$ is the discriminant of this quadratic is $<0$. (The coefficient of y^2 is positive).

If I am not wrong the discriminant is equal to $\Delta =n^2-4N^2(1-N)^2$. Since $n^2<\frac{1}{2\pi}$, the inequality will be true for $4N^2(1-N)^2>1/2\pi$.

This gives, that the inequality is true except when when $0.275214<N<0.724786$. This correspond to $|x|<0.597$.

Since the derivative of your function is easily bounded, to prove the inequality in the interval $|x|<0.597$ you may use the maximal slope principle as described in this paper http://front.math.ucdavis.edu/1004.0469

We may see that the inequality is true for $|x|>0.597$ in the following way:

For a given value of $x$ consider the values of $N$ and $n$. The inequality will be true for this $x$ if the quadratic polynomial in $y$ $$(y^2+1)N+y\, n-(y N+n)^2-N^2 >0$$ is always positive. In other words the inequality is true for this $x$ is the discriminant of this quadratic is $<0$. (The coefficient of y^2 is positive).

If I am not wrong the discriminant is equal to $\Delta =n^2-4N^2(1-N)^2$. Since $n^2<\frac{1}{2\pi}$, the inequality will be true for $4N^2(1-N)^2>1/2\pi$.

This gives, that the inequality is true except when $0.275214<N<0.724786$. This correspond to $|x|<0.597$.

Since the derivative of your function is easily bounded, to prove the inequality in the interval $|x|<0.597$ you may use the maximal slope principle as described in this paper http://front.math.ucdavis.edu/1004.0469

We may see that the inequality is true for $|x|<0.597$ in the following way:

For a given value of $x$ consider the values of $N$ and $n$. The inequality will be true for this $x$ if the quadratic polynomial in $y$ $$(y^2+1)N+y\, n-(y N+n)^2-N^2 >0$$ is always positive. In other words the inequality is true for this $x$ is the discriminant of this quadratic is $<0$. (The coefficient of y^2 is positive).

If I am not wrong the discriminant is equal to $\Delta =n^2-4N^2(1-N)^2$. Since $n^2<\frac{1}{2\pi}$, the inequality will be true for $4N^2(1-N)^2>1/2\pi$.

This gives, that the inequality is true when $0.275214<N<0.724786$. This correspond to $|x|<0.597$.

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juan
juan
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We may see that the inequality is true for $|x|<0.597$$|x|>0.597$ in the following way:

For a given value of $x$ consider the values of $N$ and $n$. The inequality will be true for this $x$ if the quadratic polynomial in $y$ $$(y^2+1)N+y\, n-(y N+n)^2-N^2 >0$$ is always positive. In other words the inequality is true for this $x$ is the discriminant of this quadratic is $<0$. (The coefficient of y^2 is positive).

If I am not wrong the discriminant is equal to $\Delta =n^2-4N^2(1-N)^2$. Since $n^2<\frac{1}{2\pi}$, the inequality will be true for $4N^2(1-N)^2>1/2\pi$.

This gives, that the inequality is true except when $0.275214<N<0.724786$. This correspond to $|x|<0.597$.

Since the derivative of your function is easily bounded, to prove the inequality in the interval $|x|<0.597$ you may use the maximal slope principle as described in this paper http://front.math.ucdavis.edu/1004.0469

We may see that the inequality is true for $|x|<0.597$ in the following way:

For a given value of $x$ consider the values of $N$ and $n$. The inequality will be true for this $x$ if the quadratic polynomial in $y$ $$(y^2+1)N+y\, n-(y N+n)^2-N^2 >0$$ is always positive. In other words the inequality is true for this $x$ is the discriminant of this quadratic is $<0$. (The coefficient of y^2 is positive).

If I am not wrong the discriminant is equal to $\Delta =n^2-4N^2(1-N)^2$. Since $n^2<\frac{1}{2\pi}$, the inequality will be true for $4N^2(1-N)^2>1/2\pi$.

This gives, that the inequality is true except when $0.275214<N<0.724786$. This correspond to $|x|<0.597$.

Since the derivative of your function is easily bounded, to prove the inequality in the interval $|x|<0.597$ you may use the maximal slope principle as described in this paper http://front.math.ucdavis.edu/1004.0469

We may see that the inequality is true for $|x|>0.597$ in the following way:

For a given value of $x$ consider the values of $N$ and $n$. The inequality will be true for this $x$ if the quadratic polynomial in $y$ $$(y^2+1)N+y\, n-(y N+n)^2-N^2 >0$$ is always positive. In other words the inequality is true for this $x$ is the discriminant of this quadratic is $<0$. (The coefficient of y^2 is positive).

If I am not wrong the discriminant is equal to $\Delta =n^2-4N^2(1-N)^2$. Since $n^2<\frac{1}{2\pi}$, the inequality will be true for $4N^2(1-N)^2>1/2\pi$.

This gives, that the inequality is true except when $0.275214<N<0.724786$. This correspond to $|x|<0.597$.

Since the derivative of your function is easily bounded, to prove the inequality in the interval $|x|<0.597$ you may use the maximal slope principle as described in this paper http://front.math.ucdavis.edu/1004.0469

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juan
juan
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