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For Problem 2:
Given that a and b are positive numbers, one can rewrite the inequality into 1/(a+b) is less than (a+b)/(ab). This implies that (a+b)^2 = a^2 + 2ab + b^2 > ab. Furthermore, a^2 + ab + b^2 > 0. Since a, b > 0, a^2 > 0, and b^2 > 0, thus, the inequality is true for positive integers.

For Problem 1:
a. Using the Vieta's theorem or the relationship between the coefficients of the quadratic equation, one can rewrite the equation as a + b = 1. Assuming a and b are roots of the quadratic equation x^2 - x + c = 0 where c = ab. We let a' and b" be the leading coefficient and coefficient of x respectively. By discriminants, in order to have a real solution, (b')^2-4(a')c must be greater or equal than to 0. By substitution, 1 - 4c must be greater than or equal to 0. Hence, c is less than or equal to 1/4 in which c = ab.

b. Using the implication in part a, squaring the equation a + b = 1 both sides, a^2 + 2ab + b^2 = 1. Since ab is less than or equal to 1/4, a^2 + b^2 + (1/2) is greater than or equal to 1, Hence, a^2 + b^2 is greater than or equal to 1/2.

Problem1 :a you have b^2-4ac >= 0 what are the values of a and b that you used to substitute with? It looks like you took a=1 and b=1 witch gives 1-4c but a+b=1 not 2.
Note: this problem is given to grade 10. discriminant is given in grade 11. So is there any other way to solve this problem?

Well, we substitute for $a$ to get $b(1-b)\le \frac{1}{4}\implies b^2-b+\frac{1}{4}\ge 0$. We find the vertex of the parabola; the x-value of the vertex is simply $\frac{-b}{2a}=\frac{1}{2}$. We plug this in the function to get $\left(\frac{1}{2}\right)^2-\frac{1}{2}+\frac{1}{4}\ge 0$ which is indeed true. Since the parabola opens up, there will not ever be a point on the parabola with lower y-value than at the vertex; therefore there does not exist any numbers $a,b$ with $a=1-b$ such that $ab> \frac{1}{4}$.

Oh.. wait... I think I used wrong notations because I kept a and b as roots and at the same time the coefficients of the quadratic equation. Thanks for informing... I'll edit.. ASAP...

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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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TopNewestFor Problem 2: Given that a and b are positive numbers, one can rewrite the inequality into 1/(a+b) is less than (a+b)/(ab). This implies that (a+b)^2 = a^2 + 2ab + b^2 > ab. Furthermore, a^2 + ab + b^2 > 0. Since a, b > 0, a^2 > 0, and b^2 > 0, thus, the inequality is true for positive integers.

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For Problem 1: a. Using the Vieta's theorem or the relationship between the coefficients of the quadratic equation, one can rewrite the equation as a + b = 1. Assuming a and b are roots of the quadratic equation x^2 - x + c = 0 where c = ab. We let a' and b" be the leading coefficient and coefficient of x respectively. By discriminants, in order to have a real solution, (b')^2-4(a')c must be greater or equal than to 0. By substitution, 1 - 4c must be greater than or equal to 0. Hence, c is less than or equal to 1/4 in which c = ab.

b. Using the implication in part a, squaring the equation a + b = 1 both sides, a^2 + 2ab + b^2 = 1. Since ab is less than or equal to 1/4, a^2 + b^2 + (1/2) is greater than or equal to 1, Hence, a^2 + b^2 is greater than or equal to 1/2.

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Problem1 :a you have b^2-4ac >= 0 what are the values of a and b that you used to substitute with? It looks like you took a=1 and b=1 witch gives 1-4c but a+b=1 not 2.

Note: this problem is given to grade 10. discriminant is given in grade 11. So is there any other way to solve this problem?

Log in to reply

Well, we substitute for $a$ to get $b(1-b)\le \frac{1}{4}\implies b^2-b+\frac{1}{4}\ge 0$. We find the vertex of the parabola; the x-value of the vertex is simply $\frac{-b}{2a}=\frac{1}{2}$. We plug this in the function to get $\left(\frac{1}{2}\right)^2-\frac{1}{2}+\frac{1}{4}\ge 0$ which is indeed true. Since the parabola opens up, there will not ever be a point on the parabola with lower y-value than at the vertex; therefore there does not exist any numbers $a,b$ with $a=1-b$ such that $ab> \frac{1}{4}$.

Log in to reply

Oh.. wait... I think I used wrong notations because I kept a and b as roots and at the same time the coefficients of the quadratic equation. Thanks for informing... I'll edit.. ASAP...

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