Suppose a,b,c are three real numbers, such that the quadratic equation x²-(a+b+c)x+(ab+bc+ac)=0,has roots of the form α±iβ, where α>0 β≠0 are real numbers (i=sqrt(-1)). Show that (I) the numbers a,b,c are all positive. (II) the numbers √a,√b,√c form the sides of a triangle.
I tried to look at the roots of the quadratic equation by using the quadratic formula.
[(a+b+c)±√((a+b+c)²-4(ab+bc+ac))]/2 = (a+b+c)/2 ± √D/2
Here by comparison we have α=(a+b+c)/2 and β=√D/2 Now the discrimant has to be negative . Hence we have (a+b+c)²<4(ab+bc+ac)
(a²+b²+c²)<2(ab+bc+ca) Now a+b+c must me positive since It corresponds to α in α±iβ which is given to be positive..Now ab+bc+ac is also positive since it is greater than sums of the perfect squares of 3 numbers.so a+b+c ,ab+bc+ac postive and twice of ab+bc+ac is greater than a²+b²+c². This is all where I got.. I also understood what they mean by √a,√b,√c form triangle. It means that sum of any pair of square roots must exceed the the third square root. But I don't understand how it is
To prove $a, b, c > 0$, you can assume one or two of them are not positive (it is clear that one at least is positive since $a + b + c > 0$) and try to contradict : $a^2 + b^2 + c^2 < 2 ab + 2 bc + 2 ac$.
For instance, if $a, b > 0$ and $c \leq 0$, you have : $a^2 + b^2 + c^2 < 2 ab - 2 |bc| - 2 |ac|$, which is equivalent to $(a-b)^2 + c^2 + 2 |bc| + 2 |ac| < 0$, impossible. The other possibility can be treated the same way.
For the second part, you can assume $0 < a \leq b \leq c$ for instance and all you have to prove is $\sqrt{a} + \sqrt{b} \geq \sqrt{c}$. Since everything is positive, this is equivalent to $a + b + 2 \sqrt{ab} \geq c$, and then to $4 ab \geq c^2 + a^2 + b^2 - 2 ac - 2 bc + 2 ab$... which is exactly the inequality you already have. (Note that, since it is even a strict inequality, the triangle is a "real" triangle, i.e. it is not a flat one.)