#### Question

If a,b,c are positive real numbers such that \dfrac{\sqrt{2}b-2c }{a}=1, show that b^2\geq 4ac

Question

If a,b,c are positive real numbers such that \dfrac{\sqrt{2}b-2c }{a}=1, show that b^2\geq 4ac

\dfrac{\sqrt{2}b-2c }{a}=1

Rearrange the equation

\sqrt{2}b = 2c+a

2c+a\geq 2\sqrt{2ca}

Then

\sqrt{2}b \geq 2\sqrt{2ca}

b\geq 2\sqrt{ac}

Square the inequality

b^2\geq 4ac