MHB Can You Prove This Triangle Inequality Without A Hint?

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The discussion centers on proving the triangle inequality expressed as (a+1)(y²a + z²) > x²a for any real number a, given that x, y, and z are the lengths of the sides of a triangle. Participants express varying levels of difficulty in solving the inequality, with some stating they could not arrive at a solution without hints. The conversation emphasizes the importance of understanding the properties of triangle sides in relation to the inequality. The challenge lies in demonstrating the inequality holds true under all conditions defined by the triangle's properties. Overall, the thread highlights the complexity of the problem and the necessity of guidance in tackling it.
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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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Subtle Hint:
Quadratic function property.
 
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

$(a+1)(y^2a+z^2) > x^2a$
 
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
 
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