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


4 years ago

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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