- a and b being two real numbers such that a=1-b.

a) Show that ab <= 1/4

b) Deduce that \(a^{2} + b^{2}\) >= 1/2 - a and b being two strictly positive numbers, show that:

1/(a+b) < (1/a) + (1/b)

- a and b being two real numbers such that a=1-b.

a) Show that ab <= 1/4

b) Deduce that \(a^{2} + b^{2}\) >= 1/2 - a and b being two strictly positive numbers, show that:

1/(a+b) < (1/a) + (1/b)

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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. – John Ashley Capellan · 3 years, 8 months ago

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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. – John Ashley Capellan · 3 years, 8 months ago

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Note: this problem is given to grade 10. discriminant is given in grade 11. So is there any other way to solve this problem? – Oussama Jaber · 3 years, 8 months ago

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– John Ashley Capellan · 3 years, 8 months ago

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...Log in to reply

– Daniel Liu · 3 years, 8 months ago

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