Can You Prove This Triangle Inequality Without A Hint?

[anemone]

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.

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

 

[anemone]

Subtle Hint:

Spoiler

Quadratic function property.

 

[kaliprasad]

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

Spoiler

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

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

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

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

 

[anemone]

I could not have solved without the hint

Spoiler

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

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

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!

Spoiler

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