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


4 years ago

mycroft holmes
272 Points
							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

4 years ago
mycroft holmes
272 Points
							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, $x +\frac{1}{x} \ge 2$ …........................................................................................

4 years ago
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