Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

First, to avoid misunderstandings, let's reformulate the problem: Problem. Let a, b, c be three real numbers. Prove that: (a) If $a\geq 0$, $c\geq 0$ and $4ac\geq b^2$, then for any pair of vectors $\overrightarrow{f}$ and $\overrightarrow{g}$, we have $a\cdot\overrightarrow{f}^2+b\cdot\overrightarrow{f}\cdot\overrightarrow{g}+c\cdot\overrightarrow{g}^2\geq 0$. (b) If for every pair of vectors $\overrightarrow{f}$ and $\overrightarrow{g}$, we have $a\cdot\overrightarrow{f}^2+b\cdot\overrightarrow{f}\cdot\overrightarrow{g}+c\cdot\overrightarrow{g}^2\geq 0$, then $a\geq 0$, $c\geq 0$ and $4ac\geq b^2$. Solution. (a) Since the numbers a and c are $\geq 0$, their square roots $\sqrt{a}$ and $\sqrt{c}$ are real numbers. Since $4ac\geq b^2$, we have $\sqrt{4ac}\geq\sqrt{b^2}$; in other words, $2\sqrt{ac}\geq\left|b\right|$. Since $\left|b\right|\geq b$ and $\left|b\right|\geq -b$, this yields $2\sqrt{ac}\geq b$ and $2\sqrt{ac}\geq -b$. In other words, $2\sqrt{ac}-b\geq 0$ and $2\sqrt{ac}+b\geq 0$. Now, if $\overrightarrow{f}\cdot\overrightarrow{g}\geq 0$, then we can write $a\cdot\overrightarrow{f}^2+b\cdot\overrightarrow{f}\cdot\overrightarrow{g}+c\cdot\overrightarrow{g}^2=\left(a\cdot\overrightarrow{f}^2-2\sqrt{ac}\cdot\overrightarrow{f}\cdot\overrightarrow{g}+c\cdot\overrightarrow{g}^2\right)+\left(2\sqrt{ac}+b\right)\cdot\overrightarrow{f}\cdot\overrightarrow{g}$ $=\underbrace{\left(\sqrt{a}\cdot\overrightarrow{f}-\sqrt{c}\cdot\overrightarrow{g}\right)^2}_{\text{trivially }\geq 0}+\underbrace{\left(2\sqrt{ac}+b\right)}_{\geq 0}\cdot\underbrace{\overrightarrow{f}\cdot\overrightarrow{g}}_{\geq 0}\geq 0$, and if $\overrightarrow{f}\cdot\overrightarrow{g}\leq 0$, we can write $a\cdot\overrightarrow{f}^2+b\cdot\overrightarrow{f}\cdot\overrightarrow{g}+c\cdot\overrightarrow{g}^2=\left(a\cdot\overrightarrow{f}^2+2\sqrt{ac}\cdot\overrightarrow{f}\cdot\overrightarrow{g}+c\cdot\overrightarrow{g}^2\right)-\left(2\sqrt{ac}-b\right)\cdot\overrightarrow{f}\cdot\overrightarrow{g}$ $=\underbrace{\left(\sqrt{a}\cdot\overrightarrow{f}+\sqrt{c}\cdot\overrightarrow{g}\right)^2}_{\text{trivially }\geq 0}-\underbrace{\left(2\sqrt{ac}-b\right)}_{\geq 0}\cdot\underbrace{\overrightarrow{f}\cdot\overrightarrow{g}}_{\leq 0}\geq 0$. Thus, problem (a) is solved. (b) First, let $\overrightarrow{f}=\overrightarrow{0}$ be the zero vector, and $\overrightarrow{g}$ be a vector of length 1. Then, $a\cdot\overrightarrow{f}^2+b\cdot\overrightarrow{f}\cdot\overrightarrow{g}+c\cdot\overrightarrow{g}^2=a\cdot\overrightarrow{0}^2+b\cdot\overrightarrow{0}\cdot\overrightarrow{g}+c\cdot\overrightarrow{g}^2=c\cdot\overrightarrow{g}^2=c\cdot\left|\overrightarrow{g}\right|^2=c\cdot 1^2=c$. Since $a\cdot\overrightarrow{f}^2+b\cdot\overrightarrow{f}\cdot\overrightarrow{g}+c\cdot\overrightarrow{g}^2\geq 0$, this yields $c\geq 0$. Similarly, $a\geq 0$. Now, if a = 0, then the inequality $a\cdot\overrightarrow{f}^2+b\cdot\overrightarrow{f}\cdot\overrightarrow{g}+c\cdot\overrightarrow{g}^2\geq 0$ becomes $b\cdot\overrightarrow{f}\cdot\overrightarrow{g}+c\cdot\overrightarrow{g}^2\geq 0$, or, equivalently, $\left(b\cdot\overrightarrow{f}+c\cdot\overrightarrow{g}\right)\cdot\overrightarrow{g}\geq 0$. But if the number b is nonzero, we easily see that, when the vector $\overrightarrow{g}$ stays fixed and the vector $\overrightarrow{f}$ varies, the vector $b\cdot\overrightarrow{f}+c\cdot\overrightarrow{g}$ can have an arbitrary direction, so particularly it can make an obtuse angle with the vector $\overrightarrow{g}$, thus yielding a negative scalar product $\left(b\cdot\overrightarrow{f}+c\cdot\overrightarrow{g}\right)\cdot\overrightarrow{g}$, what contradicts $\left(b\cdot\overrightarrow{f}+c\cdot\overrightarrow{g}\right)\cdot\overrightarrow{g}\geq 0$. Hence, the number b cannot be nonzero, i. e. we have b = 0, and thus we trivially have $4ac\geq b^2$ (since $4ac=0=b^2$). This was for a = 0. Similarly, we can handle the case when c = 0. Now, assume that neither a = 0 nor c = 0. Since $a\geq 0$ and $c\geq 0$, we thus have a > 0 and c > 0. Now, let $\overrightarrow{f}$ be an arbitrary vector of length 1, and let $\overrightarrow{g}=\dfrac{\sqrt{a}}{\sqrt{c}}\cdot\overrightarrow{f}$. Then, $\sqrt{c}\cdot\overrightarrow{g}=\sqrt{a}\cdot\overrightarrow{f}$, so that $\sqrt{a}\cdot\overrightarrow{f}-\sqrt{c}\cdot\overrightarrow{g}=0$, and thus $a\cdot\overrightarrow{f}^2+b\cdot\overrightarrow{f}\cdot\overrightarrow{g}+c\cdot\overrightarrow{g}^2=\left(a\cdot\overrightarrow{f}^2-2\sqrt{ac}\cdot\overrightarrow{f}\cdot\overrightarrow{g}+c\cdot\overrightarrow{g}^2\right)+\left(2\sqrt{ac}+b\right)\cdot\overrightarrow{f}\cdot\overrightarrow{g}$ $=\underbrace{\left(\sqrt{a}\cdot\overrightarrow{f}-\sqrt{c}\cdot\overrightarrow{g}\right)^2}_{=0}+\left(2\sqrt{ac}+b\right)\cdot\overrightarrow{f}\cdot\overrightarrow{g}=\left(2\sqrt{ac}+b\right)\cdot\overrightarrow{f}\cdot\overrightarrow{g}$ $=\left(2\sqrt{ac}+b\right)\cdot\overrightarrow{f}\cdot\dfrac{\sqrt{a}}{\sqrt{c}}\cdot\overrightarrow{f}=\left(2\sqrt{ac}+b\right)\cdot\dfrac{\sqrt{a}}{\sqrt{c}}\cdot\overrightarrow{f}^2=\left(2\sqrt{ac}+b\right)\cdot\dfrac{\sqrt{a}}{\sqrt{c}}$ (since $\overrightarrow{f}^2=\left|\overrightarrow{f}\right|^2=1^2=1$). Since $a\cdot\overrightarrow{f}^2+b\cdot\overrightarrow{f}\cdot\overrightarrow{g}+c\cdot\overrightarrow{g}^2\geq 0$ and a > 0 and c > 0, this yields $2\sqrt{ac}+b\geq 0$, so that $2\sqrt{ac}\geq -b$. Similarly, by putting $\overrightarrow{g}=-\dfrac{\sqrt{a}}{\sqrt{c}}\cdot\overrightarrow{f}$ instead of $\overrightarrow{g}=\dfrac{\sqrt{a}}{\sqrt{c}}\cdot\overrightarrow{f}$ and using $a\cdot\overrightarrow{f}^2+b\cdot\overrightarrow{f}\cdot\overrightarrow{g}+c\cdot\overrightarrow{g}^2=\underbrace{\left(\sqrt{a}\cdot\overrightarrow{f}+\sqrt{c}\cdot\overrightarrow{g}\right)^2}_{=0}-\left(2\sqrt{ac}-b\right)\cdot\overrightarrow{f}\cdot\overrightarrow{g}$, we can obtain $2\sqrt{ac}\geq b$. Now, since the absolute value $\left|b\right|$ of b is one of the numbers b and -b, this yields $2\sqrt{ac}\geq\left|b\right|$. Since both sides of this inequality are nonnegative, we can square it and obtain $4ac\geq b^2$. Thus, altogether, we have showed that $a\geq 0$, $c\geq 0$ and $4ac\geq b^2$. Hence, problem (b) is solved. Darij