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If [math]x^2-2ax+b=0[/math] ,[math]b\neq 0[/math] and has two unequal real roots then show that [math]6a^4-2a^3-2a^2b+a^2-6a^2b+2ab+2b^2-b\geq 0[/math]

I know that since the equation has two unequal real roots then [math]4a^2-4b>0[/math] but then how does this figure in the above inequality??

Thank you .

But the frame within which we work is high school level. Thus although we know a little bit of vectors we have not done the Cauchy–Schwarz inequality.

Anyway thanks for the reference i will look it up

Before i learn your proof ,there is a thing that i am worried about.

if we prove [math]\frac{x^4+y^4+z^4}{x^2y^2+x^2z^2+y^2z^2}\geq 1>\frac{3m+2n}{3(m+n)}[/math] that implies [math]\frac{x^4+y^4+z^4}{x^2y^2+x^2z^2+y^2z^2}>\frac{3m+2n}{3(m+n)}[/math] and not

[math]\frac{x^4+y^4+z^4}{x^2y^2+x^2z^2+y^2z^2}\geq\frac{3m+2n}{3(m+n)}[/math] .

Is that right??

Find the sign of the following polynomial (i.e whether is greater or less than zero)

[math]x^4+x^3(a+b)+x^2(a+ab+b)+x(a^2+b^2) +ab[/math],given that

[math]|a|<\sqrt{b}[/math] and [math]|b|<\sqrt{a}[/math]

Since [math] x^4 ,x^2[/math] are positive if we can show that (a+ab+b)>0,ab>0 and also [math]x^3(a+b)-x(a^2+b^2)>0[/math] are we done??

Thanks .

In that case because 2m<3m ,then 2m+3n<3(m+n) or [math]\frac{2m+3n}{3(m+n)}<1\leq\frac{x^4+y^4+z^4}{x^2y^2+x^2z^2+y^2z^2}[/math].

I am worried about the equality sign, because [math]A<1\leq B[/math] =>A<B.

Is it easy to prove [math]\frac{x^4+y^4+z^4}{x^2y^2+x^2z^2+y^2z^2}\geq 1[/math]??

Thank you all ,but could you be more specific?

R.H.S is a Natural No ,but what manipulations can make the ratio:[math]\frac{x^4+y^4+z^4}{x^2y^2+x^2z^2+y^2z^2}[/math] such a complicated Natural No??.

Why not a whole Natural No??

I have to hand in that exercise next Monday and i am still in the dark.

Thanx

Well, the equation is symmetric about A, B and C for a start, so it doesn't matter which one you decide to be equal to 90. After that you'll need to look at various trig identities that you know.

Thanks i will try in that direction

Here is another problem that made me land on my head:

if [math]sec\frac{A}{2}sec\frac{B}{2}sec\frac{C}{2}+sec\frac{A}{2}+sec\frac{B}{2}+sec\frac{C}{2} -2=0[/math] ,then one of the angles (A,B,C) of the triangle ABC is 90 degrees.

How do we start??

I was trying to prove the following inequality and really got stuck

The inequality is. [math]\frac{x^4+y^4+z^4}{x^2y^2+x^2z^2+y^2z^2}\geq\frac{2m+3n}{3(m+n)}[/math] for all the reals x,y,z different to zero and for all the natural Nos m,n

Can we use induction here