Question

Prove that $(a^2+b^2)x^2-2b(a+c)x+(b^2+c^2)\ge 0$ for all $x\in R.$ If equality holds then find the ratio of the roots of the equation $a{x}^2+2bx+c=0.$

Answer 1

Prove that:

$(a^2+b^2)x^2-2b(a+c)x+(b^2+c^2)\ge 0$

If ${(-2b(a+c))}^2-4(a^2+b^2)(b^2+c^2)\le 0$

$\Rightarrow {4b}^2{(a+c)}^2-4(a^2+b^2)(b^2+c^2)\le 0$

$\Rightarrow b^2(a^2+c^2+2ac)\le (a^2+b^2)(b^2+c^2)$

$\Rightarrow a^2{b}^2+b^2{c}^2+2b^{2}ac\le a^2{b}^2+a^2{c}^2+b^4+b^2{c}^2$

$\Rightarrow {2b}^{2}ac\le a^2{c}^2+b^4$

$\Rightarrow \frac{a^2{c}^2+b^4}{2}\ge b^{2}ac$

We know $AM\ge GM$

$\therefore$ $\frac{a^2{c}^2+b^4}{2}\ge \sqrt{b^4.a^2{c}^2}$

$\Rightarrow \frac{a^2{c}^2+b^4}{2}\ge b^{2}ac$

$\therefore$ $\left( { a }^{ 2 }+{ b }^{ 2 } \right) { x }^{ 2 }-2b\left( a+c \right)

x+{ b }^{ 2 }+{ c }^{ 2 }\ge 0$

Hence, proved