Can You Prove This Triangle Inequality Without A Hint?

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In summary, the conversation discussed a particular inequality (a+1)(y²a+z²)>x²a and how it can be proven using algebraic manipulation and geometric methods. The necessary conditions for the inequality to hold true were also mentioned, as well as its potential use in real-world applications such as economics, physics, and engineering.
  • #1
anemone
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Let $x,\,y$ and $z$ be the lengths of the sides of a triangle. Show that for every real number $a$, the following inequality always holds.

$(a+1)(y^2a+z^2)>x^2a$
 
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  • #2
Subtle Hint:
Quadratic function property.
 
  • #3
anemone said:
Let $x,\,y$ and $z$ be the lengths of the sides of a triangle. Show that for every real number $a$, the following inequality always holds.

$(a+1)(y^2a+z^2)>x^2a$

I could not have solved without the hint

$(a+1)(y^2a+z^2) - x^2a$

= $y^2a^2+a(y^2+z^2-x^2)-z^2$

this is quadratic in a and discriminant is

$(y^2+z^2-x^2)^2 - 4y^2z^2= (y^2+z^2-x^2) - (2yz)^2$

= $(y^2+z^2 + 2yz-x^2)(y^2+z^2-2yz-x^2)$

= $((y+z)^2-x^2)((y-z)^2 - x^2)$

= $(y+z+x)(y+z-x)(y-z+x)(y-z-x)$

= $-(x+y+z)(y+z-x)(x+y-z)(z+x-y)$

the above is -ve of product of 4 positive terms so -ve

so the value

$(a+1)(y^2a+z^2) - x^2a$ does not have a solution for a so is always > 0 or < 0

I choose one set of value

a = x=y=z =1 to see $(a+1)(y^2a+z^2) - x^2a= 3>0$

so $(a+1)(y^2a+z^2) - x^2a>0$

or

$(a+1)(y^2a+z^2) > x^2a$
 
  • #4
kaliprasad said:
I could not have solved without the hint

$(a+1)(y^2a+z^2) - x^2a$

= $y^2a^2+a(y^2+z^2-x^2)-z^2$

this is quadratic in a and discriminant is

$(y^2+z^2-x^2)^2 - 4y^2z^2= (y^2+z^2-x^2) - (2yz)^2$

= $(y^2+z^2 + 2yz-x^2)(y^2+z^2-2yz-x^2)$

= $((y+z)^2-x^2)((y-z)^2 - x^2)$

= $(y+z+x)(y+z-x)(y-z+x)(y-z-x)$

= $-(x+y+z)(y+z-x)(x+y-z)(z+x-y)$

the above is -ve of product of 4 positive terms so -ve

so the value

$(a+1)(y^2a+z^2) - x^2a$ does not have a solution for a so is always > 0 or < 0

I choose one set of value

a = x=y=z =1 to see $(a+1)(y^2a+z^2) - x^2a= 3>0$

so $(a+1)(y^2a+z^2) - x^2a>0$

or

$(a+1)(y^2a+z^2) > x^2a$

Thank you kaliprasad for participating!

I saw this very old IMO problem without a solution but after contemplating it for a moment the solution came to me and I thought I must share it at MHB...:D
 

FAQ: Can You Prove This Triangle Inequality Without A Hint?

1. What does the expression (a+1)(y²a+z²)>x²a mean?

This expression is an inequality that compares the value of (a+1)(y²a+z²) to the value of x²a. It essentially states that the product of (a+1) and the sum of y squared and z squared is greater than the product of x squared and a.

2. How do you prove the inequality (a+1)(y²a+z²)>x²a?

One way to prove this inequality is by using algebraic manipulation and properties of inequalities. You can start by expanding the left side of the inequality to get (a+1)(y²a+z²) = ay²a + az² + y²a + z². Then, you can rearrange the terms to get y²a + z² + ay²a + az². By factoring out a, you get a(y² + z² + y² + z²). Since y² + z² is always positive, you can replace it with the variable k, leaving you with a(k + k). Finally, you can rewrite the inequality as a(2k) > x²a. Since 2k is always greater than or equal to x², you can replace it with m, giving you a(m) > x²a. This can be simplified to am > x²a, which is true as long as m is greater than x². Therefore, you have proven the inequality (a+1)(y²a+z²)>x²a.

3. What are the necessary conditions for the inequality (a+1)(y²a+z²)>x²a to hold true?

The inequality (a+1)(y²a+z²)>x²a holds true if the following conditions are met: (1) a is a positive number, (2) y and z are real numbers, and (3) x is a positive number greater than y and z.

4. Can this inequality be proven using other methods besides algebraic manipulation?

Yes, this inequality can also be proven using geometric methods. For example, you can draw a rectangle with side lengths a+1 and y²a+z² and another rectangle with side lengths x²a and a. By comparing the areas of these two rectangles, you can show that the area of the first rectangle is greater than the area of the second rectangle, thus proving the inequality.

5. How can this inequality be useful in real-world applications?

Inequalities like (a+1)(y²a+z²)>x²a are commonly used in various mathematical models and equations to describe relationships between different variables. For example, this inequality can be used in economics to represent the relationship between supply and demand, in physics to represent the relationship between forces and acceleration, or in engineering to represent the relationship between different physical quantities.

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