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Grade: 11
        
if (x^5)-(x^3)+ x =a,x>0,then prove that   (x^6)>=2a-1
3 years ago

Answers : (2)

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
3 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
 
…........................................................................................
3 years ago
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