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Grade 12Algebra

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

Question image for Q3,4..............................................

Aviral alok

9 Years agoGrade 12

2 Answers

mycroft holmes

9 Years ago

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

mycroft holmes

9 Years ago

$x^4+\frac{1}{x^4}, x^3+\frac{1}{x^3} \in \mathbb{Q} \Rightarrow x^8+\frac{1}{x^8}, x^6+\frac{1}{x^6} \in \mathbb{Q}$

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 $x+\frac{1}{x} \in \mathbb{Q}$

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