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Grade: 12

                        

Q3,4............................................................

4 years ago

Answers : (2)

mycroft holmes
272 Points
							
By Cauchy Schwarz Ineq you have
 
LHS \ge \frac{(a+b+c+d+e)^2}{a+b+c+d+e-5}
Let a+b+c+d+e = t.
 
Then we need to prove that t2/(t-5) \ge 20 which is equivalent to ptoving that
 
t2-20t+100 = (t-10)2 \ge 0 which is true as any perfect square is non-negative
4 years ago
mycroft holmes
272 Points
							
x^4+\frac{1}{x^4}, x^3+\frac{1}{x^3} \in \mathbb{Q} \Rightarrow x^8+\frac{1}{x^8}, x^6+\frac{1}{x^6} \in \mathbb{Q}
 
Further we have x^8+\frac{1}{x^8} = \left(x^6+\frac{1}{x^6} \right)\left(x^2+\frac{1}{x^2} \right) - \left(x^4+\frac{1}{x^4} \right)
 
and hence we must have x^2+\frac{1}{x^2} \in \mathbb{Q}
 
Finally
 
x^3+\frac{1}{x^3} = \left (x^2+\frac{1}{x^2} \right ) \left (x+\frac{1}{x} \right ) - \left (x+\frac{1}{x} \right )
 
\Rightarrow x^3+\frac{1}{x^3} = \left (x+\frac{1}{x} \right ) \left (x^2+\frac{1}{x^2} -1 \right)
 
from which we immediately have x+\frac{1}{x} \in \mathbb{Q}
…........................................................................................
4 years ago
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