Mark Your Own STEP Attempt

For each line below, mark yourself twice: R if you reached it, and J if your write-up justifies it well enough that a marker would not have to fill anything in. An R without a J is the characteristic STEP loss.

(i) [4 marks]

• The identity $\dfrac{p^2}{x}+\dfrac{q^2}{y}-\dfrac{(p+q)^2}{x+y}=\dfrac{(py-qx)^2}{xy(x+y)}$, valid for all real $p$ and $q$ and all positive $x$ and $y$.
• $xy(x+y) \gt 0$ and $(py-qx)^2 \ge 0$, hence $\dfrac{p^2}{x}+\dfrac{q^2}{y} \ge \dfrac{(p+q)^2}{x+y}$.
• Equality holds if and only if $py=qx$, equivalently $\dfrac{p}{x}=\dfrac{q}{y}$, and both directions are established.
• 2 marks: Combining the three fractions over $xy(x+y)$ and simplifying the numerator to $(py-qx)^2$ or to an equivalent perfect square. One of these two marks is for the correct expansion; the other is for recognising the numerator as a square rather than leaving it as $p^2y^2-2pqxy+q^2x^2$.
• 1 marks: Deducing the inequality, with the positivity of the denominator stated. A solution that multiplies through by $xy(x+y)$ instead earns this mark only if it says that this quantity is positive.
• 1 marks: The equality condition $py=qx$, argued in both directions: that $py=qx$ gives equality, and that equality forces $py=qx$ because the denominator cannot vanish.
• No credit: Multiplying the inequality through by $xy(x+y)$ without stating that this quantity is positive. The direction of the resulting inequality is then unjustified, and this is the standard way to lose the deduction mark.
• No credit: Quoting the Cauchy-Schwarz inequality, or naming this result as a known lemma, in place of a proof. The question asks for the difference in its simplest form, so the algebra is the required work.
• No credit: Verifying the inequality at particular values of $p,q,x,y$, or arguing from a sketch. Neither establishes a statement quantified over all real $p$ and $q$.

(ii) [6 marks]

• A stated proposition $S(n)$ quantified over all real $p_1,\dots,p_n$ and all positive $x_1,\dots,x_n$, carrying the equality criterion as part of the statement.
• The base case: $S(2)$ is part (i).
• The inductive step inequality $\sum_{i=1}^{n+1}\frac{p_i^2}{x_i} \ge \frac{P_n^2}{X_n}+\frac{p_{n+1}^2}{x_{n+1}} \ge \frac{P_{n+1}^2}{X_{n+1}}$, with the first inequality attributed to $S(n)$ and the second to part (i) applied with $p=P_n$, $q=p_{n+1}$, $x=X_n$, $y=x_{n+1}$, and with $X_n \gt 0$ noted.
• Equality in $S(n+1)$ holds if and only if it holds at both inequalities of the chain.
• If $\frac{p_1}{x_1}=\cdots=\frac{p_{n+1}}{x_{n+1}}=k$ then $P_n=kX_n$, so both inequalities are equalities.
• Conversely, equality forces $\frac{p_1}{x_1}=\cdots=\frac{p_n}{x_n}=k$ by $S(n)$, hence $\frac{P_n}{X_n}=k$, and then part (i) forces $\frac{p_{n+1}}{x_{n+1}}=\frac{P_n}{X_n}=k$.
• 1 marks: A correctly stated proposition, quantified over all admissible lists of length $n$ and including the equality clause, together with the base case identified as part (i).
• 2 marks: The two-step chain of the inductive step, with each inequality attributed to the result that justifies it and with the positivity of $X_n$ stated where part (i) is invoked.
• 1 marks: The direction: equal ratios imply equality. The mark requires the intermediate observation $P_n=kX_n$, not just an assertion.
• 2 marks: The converse direction: equality implies equal ratios. One mark for using the equality clause of the inductive hypothesis to get the first $n$ ratios equal, one for the deduction $\frac{p_{n+1}}{x_{n+1}}=\frac{P_n}{X_n}=k$ from the equality case of part (i).
• No credit: Proving the inequality and asserting the equality criterion. The criterion carries half the marks in this part, and the converse direction is the half that is not routine.
• No credit: An inductive hypothesis stated for one fixed list of $n$ numbers. The step applies the hypothesis to the first $n$ entries of a longer list, so the hypothesis has to be quantified over all lists of that length.
• No credit: An inductive step written as a chain of equivalences starting from the statement to be proved and ending at something true, with no remark that the steps are reversible.
• No credit: Omitting the base case, or checking it at $n=1$ while claiming the equality criterion holds there in the stated form: with one term the chain of ratios has a single entry and asserts nothing.

(iii) [4 marks]

• The choice $p_i=a_i$, $x_i=a_{i+1}$ with $a_{n+1}=a_1$, and the observation that $x_1+\cdots+x_n=a_1+\cdots+a_n$ because the denominators are a rearrangement of the numerators' bases.
• Part (ii) then gives $\sum_{i=1}^{n}\frac{a_i^2}{a_{i+1}} \ge \frac{(a_1+\cdots+a_n)^2}{a_1+\cdots+a_n}=a_1+\cdots+a_n$.
• Equality holds if and only if $\frac{a_1}{a_2}=\frac{a_2}{a_3}=\cdots=\frac{a_n}{a_1}$.
• Writing that common value as $k$, the product of the $n$ equal fractions is $1$, so $k^n=1$; since $k \gt 0$ this gives $k=1$ and hence $a_1=\cdots=a_n$. The converse is checked.
• 1 marks: Choosing the cyclic weights and stating why the sum of the denominators equals the sum of the $a_i$.
• 1 marks: The resulting inequality, correctly quoted from part (ii).
• 2 marks: The equality argument. One mark for reducing to the equal-ratio condition and forming the product of the $n$ ratios; one for $k^n=1$ with $k \gt 0$ giving $k=1$, and for checking the converse.
• No credit: Deducing $a_1=\cdots=a_n$ directly from the equal-ratio condition without using the ratio that closes the cycle. Equality of $\frac{a_1}{a_2},\dots,\frac{a_{n-1}}{a_n}$ alone says only that the $a_i$ form a geometric progression, which does not force them to be equal.
• No credit: Taking $k^n=1$ to give $k=1$ without noting that $k$ is positive. Over the reals with $n$ even, $k=-1$ also satisfies $k^n=1$.
• No credit: Applying part (ii) with $x_i=a_i$, which yields $a_1+\cdots+a_n$ on both sides and proves nothing about the stated sum.

(iv) [6 marks]

• $S-a_i \gt 0$ for every $i$, because $n \ge 2$ and every $a_j$ is positive.
• The construction $\frac{a_i}{S-a_i}=\frac{a_i^2}{a_i(S-a_i)}$ with $a_i(S-a_i) \gt 0$.
• $\sum_{i=1}^{n}a_i(S-a_i)=S^2-Q$ where $Q=a_1^2+\cdots+a_n^2$, and $S^2-Q \gt 0$.
• By part (ii), $\sum_{i=1}^{n}\frac{a_i}{S-a_i} \ge \frac{S^2}{S^2-Q}$.
• $Q \ge \frac{S^2}{n}$, obtained from part (ii) with $p_i=a_i$ and $x_i=1$.
• $S^2-Q \le \frac{(n-1)S^2}{n}$, both sides positive, hence $\frac{S^2}{S^2-Q} \ge \frac{n}{n-1}$.
• Equality holds if and only if $a_1=a_2=\cdots=a_n$, established in both directions.
• 2 marks: The construction that puts each term in the form (square) over (positive), and the application of part (ii) to it. One mark for the construction, one for the correct resulting bound.
• 1 marks: Evaluating $\sum_i a_i(S-a_i)$ as $S^2-Q$ and stating that this is positive.
• 2 marks: The lower bound $Q \ge \frac{S^2}{n}$ and the correctly directed deduction to $\frac{n}{n-1}$. One mark for obtaining the bound on $Q$, one for turning a lower bound on $Q$ into a lower bound on $\frac{S^2}{S^2-Q}$ with the reversal explained.
• 1 marks: Equality precisely when all the $a_i$ are equal, with both the forward deduction from the two equality criteria and the verification that equal values give $\frac{n}{n-1}$.
• No credit: Writing $\frac{a_i}{S-a_i}=\frac{(\sqrt{a_i})^2}{S-a_i}$ and applying part (ii) to reach $\frac{(\sqrt{a_1}+\cdots+\sqrt{a_n})^2}{(n-1)S}$. The step is valid but the bound is too weak to finish: closing it would need $(\sum\sqrt{a_i})^2 \ge nS$, and part (ii) with $p_i=\sqrt{a_i}$, $x_i=1$ gives the opposite inequality. Work that stops here scores nothing beyond the positivity of the denominators.
• No credit: Bounding $Q$ above rather than below. An upper bound for $Q$ gives an upper bound for $\frac{S^2}{S^2-Q}$, which is the wrong direction and settles nothing.
• No credit: Assuming $S=1$ without saying why that is permitted. The claim is homogeneous of degree $0$ in the $a_i$, so the normalisation is legitimate, but the sentence justifying it has to be there.
• No credit: Verifying the inequality only at $a_1=\cdots=a_n$ and concluding that this is the minimum.