[tex](ax+by)^2 \leq ax^2+by^2[/tex] for every a,b which satisfy

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In summary, the inequality (ax+by)^2 ≤ ax^2+by^2 holds true for any values of a and b that satisfy 0 ≤ a,b ≤ 1 and a+b=1. To prove this, we can expand the left side and simplify, using the fact that b=1-a. This will result in (a-a^2)(x-y)^2+y^2[2ab] ≥ 0. By using the fact that a+b=1, we can simplify further to get (a-a^2)[(x-y)^2-y^2]+(b-b^2)y^2. This proves the inequality and shows that there are more efficient approaches to solving such problems.
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[tex](ax+by)^2 \leq ax^2+by^2[/tex] for every a,b which satisfy [tex]0\leq a,b\leq1[/tex] and a+b=1.

My book consider this as trivial however I have hard time to prove this, will appreciate your help.
 
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  • #2


Expand the left side, let b=1-a and simplify. See if you can convert all factors involving a into a2-a :wink:
 
  • #3


I have tried many things including this one, how should I approach these kind of problems?
What is the intuition? Is the solution process of these kind of problems really involves trial and error method or I miss something?
 
  • #4


Eventually i came to this nasty prove:

[tex]ax^2+by^2-(ax+by)^2=ax^2+by^2-a^2x^2-2abxy-b^2y^2=(a-a^2)x^2-2abxy+(b-b^2)y^2=(a-a^2)x^2-2(a-a^2)xy+(b-b^2)y^2=(a-a^2)[(x-y)^2-y^2]+(b-b^2)y^2=
(a-a^2)(x-y)^2+y^2[b-b^2+a^2-a]=(a-a^2)(x-y)^2+y^2[2ab] \geq 0[/tex]

Is there a more efficient approaches to such problems?
 
  • #5


Yes, use the fact that a+b=1, thus b=1-a. Use this conversion early on.
 

Related to [tex](ax+by)^2 \leq ax^2+by^2[/tex] for every a,b which satisfy

1. What is the meaning of [tex](ax+by)^2 \leq ax^2+by^2[/tex]?

The equation [tex](ax+by)^2 \leq ax^2+by^2[/tex] represents a mathematical inequality, where the square of the sum of two terms (ax and by) is less than or equal to the sum of the squares of those terms (ax^2 and by^2). This inequality holds true for all values of a and b that satisfy the equation.

2. How is this equation related to the Pythagorean theorem?

The equation [tex](ax+by)^2 \leq ax^2+by^2[/tex] is related to the Pythagorean theorem in the sense that it is a special case of the theorem. In the Pythagorean theorem, the sum of the squares of the sides of a right triangle is equal to the square of the hypotenuse. In this equation, the sum of the squares of two terms (ax and by) is less than or equal to the square of the sum of those terms, which is similar to the Pythagorean theorem.

3. What are the solutions to this equation?

The solutions to the equation [tex](ax+by)^2 \leq ax^2+by^2[/tex] are all the values of a and b that make the inequality true. This includes an infinite number of solutions, as there are infinite possible combinations of a and b that can satisfy the equation. Some specific solutions include a=0 and b=0, a=1 and b=1, and a=-1 and b=1.

4. Can this equation be simplified?

Yes, this equation can be simplified by expanding the squared terms and rearranging them. This results in [tex]a^2x^2+2abxy+b^2y^2 \leq ax^2+by^2[/tex]. However, this may not always be necessary or helpful, as the original equation is already in a simplified form.

5. How is this equation used in real-life applications?

The equation [tex](ax+by)^2 \leq ax^2+by^2[/tex] has various real-life applications, such as in physics, engineering, and economics. In physics, it can be used to calculate the potential energy of an object in a gravitational field. In engineering, it can be used to design structures that can withstand certain forces. In economics, it can be used to model production possibilities and resource allocation. Overall, this equation helps in understanding and solving various real-life problems involving inequalities and relationships between different variables.

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