IIT JEE , AIEEE Forum - Mathematics , Algebra

JimboJones

Let

Note $a \ne 0$ as the equation becomes and thus has real roots. Similarly $b \neq 0$ and $c \neq 0$

If $\alpha + i \beta$ is one of the roots of the then the other root is the complex conjugate, $\alpha - i \beta$

Thus the sum of roots is $2\alpha = a + b + c$

$\alpha > 0$ implies

Implies one of is

Let

Now,

$(x - a) (x -b) (x -c) & = & x^3 - (a + b + c)x^2 + (ab + bc + ca)x -abc \\& = & x (x^2 - (a + b + c)x + (ab + bc + ca)) - abc \\& = & x F(x) - abc \\$

Now, for all . Why?. If for some , then we can find large enough such that . This implies that has a real root.\\

Therefore, . Since and ,it implies .

Thus, either and ,or and .

Lets assume, and

Since the roots of are complex, $\Delta < 0$

$\Delta & = & (a + b + c)^2 - 4(ab + bc + ca) \\& = & a^2 + b^2 + c^2 - 2ab - 2bc - 2ac \\& = & a^2 + (b - c)^2 - 2a(b + c)$ .

Since, , and ,

Thus $\Delta > 0$ . A contradiction.

Thus, , and

For the second part, it suffices to show that the $\sqrt{a}$ , $\sqrt{b}$ , $\sqrt{c}$ , satisfies the triangle inequality

i.e

$\sqrt{a} + \sqrt{b} > \sqrt{c}$

Squaring both sides.

$a + b + 2\sqrt{a}\sqrt{b} > c \\2\sqrt{a}\sqrt{b} > c - (a + b) \\4ab > (c - (a + b))^2 \\4ab > c^2 + a^2 + b^2 + 2ab - 2ac - 2bc \\0 > a^2 + b^2 + c^2 - 2ab - 2bc - 2c \\0 > \Delta \\$

Since $\Delta$ is a symmetric equation.

$\sqrt{b} + \sqrt{c} > \sqrt{a} \\\sqrt{c} + \sqrt{a} > \sqrt{b}$

Thus, $\sqrt{a}$ , $\sqrt{b}$ , $\sqrt{c}$ , are sides of a triangle.

This completes the proof.

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