Standard +0.3 This is a straightforward proof by contradiction applying AM-GM inequality. Students assume the negation, manipulate to reach a contradiction via the standard inequality (a+b)/2 ≥ √(ab), requiring only 2-3 algebraic steps. The technique is routine for P4/Further Pure, making it slightly easier than average.
As \(x\) and \(y\) are positive real numbers, this is a contradiction and so \(\dfrac{9x}{y}+\dfrac{y}{x}<6\) must be incorrect and so \(\dfrac{9x}{y}+\dfrac{y}{x}\geq6\)
A1*
Makes a suitable conclusion
## Question 8:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Assume there exist positive real numbers $x$ and $y$ such that $\dfrac{9x}{y}+\dfrac{y}{x}<6$ | B1 | Starts the proof by contradicting the given statement |
| $\dfrac{9x}{y}+\dfrac{y}{x}<6\Rightarrow9x^2+y^2<6xy$ (as $x$ and $y$ are both positive) | M1 | Multiplies through by $xy$ |
| $\Rightarrow9x^2+y^2-6xy<0\Rightarrow(3x-y)^2<0$ | A1 | Reaches a correct contradictory statement |
| As $x$ and $y$ are positive real numbers, this is a contradiction and so $\dfrac{9x}{y}+\dfrac{y}{x}<6$ must be incorrect and so $\dfrac{9x}{y}+\dfrac{y}{x}\geq6$ | A1* | Makes a suitable conclusion |
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8. Use proof by contradiction to prove that, for all positive real numbers $x$ and $y$,
$$\frac { 9 x } { y } + \frac { y } { x } \geqslant 6$$
\hfill \mbox{\textit{Edexcel P4 2022 Q8 [4]}}