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Grade 11Algebra

if (x^5)-(x^3)+ x =a,x>0,then prove that (x^6)>=2a-1

kriti

9 Years agoGrade 11

2 Answers

mycroft holmes

9 Years ago

Equivalently we have to prove that $x^6 +1 \ge 2(x^5-x^3+1)$ or

$x^6 +2x^3+1 \ge 2(x^5+1)$ and dividing by x³ on both sides, we have

$x^3 + \frac{1}{x^3} +2\ge 2\left(x^2+\frac{1}{x^2} \right)$

Setting $y = x+\frac{1}{x}$ this is equivalent to the inequality

$y^3 -3y+2 \ge 2y^2-4$ which is same as $y^3 -2y^2-3y+6 \ge 0$ or

$(y-2)(y^2-3) \ge 0$

From A.M – G.M Inequality, we have $y = x+\frac{1}{x} \ge 2 \Rightarrow y^2 \ge 4>3$

Hence we do have $(y-2)(y^2-3) \ge 0$ proving the inequality. Equality occurs when x=1

mycroft holmes

9 Years ago

Later realized there is a far simpler way:

We are given $x(x^4-x^2+1) =a$

So, $x^6 +1 = (x^2+1)(x^4-x^2+1) =\left( \frac{x^2+1}{x} \right ) \ a$

$= \left( x +\frac{}1{x} \right ) \ a \ge 2a$

as by AM-GM, $x +\frac{1}{x} \ge 2$

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