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 $$b^2\geq 4ac   $$

Uniteasy Admin · Accepted Answer

$$\dfrac{\sqrt{2}b-2c }{a}=1$$
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
$$b^2\geq 4ac $$

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