2022TrialShort answer

Q13 (c) (iv)

3 marks

Question

Hence prove by induction for all $a_1,a_2,a_3,\ldots,a_n>0$ that
$(a_1+a_2+\cdots+a_n)\left(\frac1{a_1}+\frac1{a_2}+\cdots+\frac1{a_n}\right)\ge n^2.$

(Note: this has already been proven for $n=1$ and $n=2$ .)

Try the question before revealing the answer.

Solution

(iv) Assume true for $n=k$ ,

i.e.

(a_1+a_2+\cdots+a_k)\left(\frac1{a_1}+\frac1{a_2}+\cdots+\frac1{a_k}\right)\ge k^2.

Prove true for $n=k+1$

i.e. RTP

(a_1+a_2+\cdots+a_k+a_{k+1})\left(\frac1{a_1}+\frac1{a_2}+\cdots+\frac1{a_k}+\frac1{a_{k+1}}\right)\ge (k+1)^2.

LHS

[(a_1+a_2+\cdots+a_k)+a_{k+1}]\left[\left(\frac1{a_1}+\frac1{a_2}+\cdots+\frac1{a_k}\right)+\frac1{a_{k+1}}\right]

=(a_1+a_2+\cdots+a_k)\left(\frac1{a_1}+\frac1{a_2}+\cdots+\frac1{a_k}\right) +\left(\frac{a_1}{a_{k+1}}+\frac{a_2}{a_{k+1}}+\cdots+\frac{a_k}{a_{k+1}}+\frac{a_{k+1}}{a_1}+\cdots+\frac{a_{k+1}}{a_k}\right)+1

=(a_1+a_2+\cdots+a_k)\left(\frac1{a_1}+\frac1{a_2}+\cdots+\frac1{a_k}\right) +\left[\frac{a_1}{a_{k+1}}+\frac{a_{k+1}}{a_1}+\cdots+\frac{a_k}{a_{k+1}}+\frac{a_{k+1}}{a_k}\right]+1

\ge k^2+\left[2+2+\cdots+2\right]+1 \quad\text{(by assumption) (using (ii))}

=k^2+2k+1

=(k+1)^2

=\text{RHS. QED. True by the principle of mathematical induction.}

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Proof

Further Proof by Mathematical InductionUse mathematical induction for series, divisibility, inequalities, algebra, calculus, probability, geometry and recursive formulae.

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