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The question original text is...

Given that $a, b, c$ are distinct real numbers such that expressions $ax^2+bx+c, bx^2+cx+a$ and $cx^2+ax+b$ are always non-negative. Prove that the quantity ${a^2+b^2+c^2 \over ab+bc+ca}$ can never lie in $(- \infty, 1] \cup [4, \infty).$

So, basically we are supposed to prove that value of ${a^2+b^2+c^2 \over ab+bc+ca}$ lies in $(1,4)$. I solved solved that ${a^2+b^2+c^2 \over ab+bc+ca}<4$ but cannot solve for minimum value.

Quadratic expressions $ax^2+bx+c$, $bx^2+cx+a$, $cx^2+ax+b$ are non negative, so $a>0,\ b>0,\ c>0$ and for discriminant, $$b^2-4ac \le 0 \\ c^2-4ba \le 0 \\ a^2-4cb \le 0$$ If all of the discriminants were $0$, then we will have $b^2=4ac$ and $c^2=4ba$, putting value of $a$ from former to latter we will have $b^3=c^3$, since both $b$ and $c$ are positive $b=c$, similarly from other equations we will finally get $$a=b=c$$ which will throw a contradiction since $b^2=4ac$. So, all three discriminants cannot be simultaneously $0$. Hence, sum of all three discrimininats will be less than $0$, i.e. $$ a^2+b^2+c^2-4(ab+bc+ca) < 0$$ From this we will get $${a^2+b^2+c^2 \over ab+bc+ca} < 4$$

I am not able to prove that ${a^2+b^2+c^2 \over ab+bc+ca} > 1$

knoftrix
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1 Answers1

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We can say: $$ac=\frac{b^2+\xi_1}{4} \land ab=\frac{c^2+\xi_2}{4} \land bc=\frac{a^2+\xi_3}{4}, \xi_1, \xi_2,\xi_3\geq0$$ We have: $$\frac{4(a^2+b^2+c^2)}{a^2+b^2+c^2+\xi_1+\xi_2+\xi_3}$$ So, when $\xi_1=\xi_2=\xi_3=0$, the fraction is equal to $4$ and it can be greater because in that case $\xi_1+\xi_2+\xi_3<0$ that is a contradiction.

Now, we suppose that: $$\frac{4(a^2+b^2+c^2)}{a^2+b^2+c^2+\xi_1+\xi_2+\xi_3}<1$$holds for every $a,b,c$. This implies: $$3(a^2+b^2+c^2)<\xi_1+\xi_2+\xi_3$$ Substituing, we have: $$4(a^2+b^2+c^2)<4(ab+ac+bc)$$ that is uncorrect because we know that: $$a^2+b^2+c^2<4(ab+ac+cb)$$

Matteo
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