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Proof, EXT2 P1 2025 HSC 14d

Positive real numbers \(a, b, c\) and \(d\) are chosen such that  \(\dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}\)  and  \(\dfrac{1}{d}\)  are consecutive terms in an arithmetic sequence with common difference \(k\), where  \(k \in \mathbb{R} , k>0\).

Show that  \(b+c<a+d\).   (3 marks)

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\(\text{AP:} \ \dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}, \dfrac{1}{d} \ \ \text{where common difference}=k\ \ (k>0)\)

\(\text{Show} \ \ b+c<a+d\)

\(\dfrac{1}{a}<\dfrac{1}{a}+k(k>0)\)

\(\dfrac{1}{a}<\dfrac{1}{b}\)

\(\text{Similarly for}\ \ \dfrac{1}{b}<\dfrac{1}{c}\  \ldots\  \text{such that}\)

\(\dfrac{1}{a}<\dfrac{1}{b}<\dfrac{1}{c}<\dfrac{1}{d} \ \ \Rightarrow\ \ a>b>c>d\ \ldots\ (1)\)
 

\(\dfrac{1}{b}=\dfrac{1}{a}+k \ \ \Rightarrow \ \ k=\dfrac{1}{b}-\dfrac{1}{a}=\dfrac{a-b}{a b}\ \ \Rightarrow \ \ a-b=kab\)

\(\text {Similarly:}\)

\(\dfrac{1}{d}=\dfrac{1}{c}+k \ \ \Rightarrow \ \ k=\dfrac{c-d}{cd}\ \ \Rightarrow\ \ c-d=kcd\)
 

\(\text{Using \(\ a>b>c>d\ \)  (see (1) above):}\)

\(a-b>c-d\)

\(b+c<a+d\)

Show Worked Solution

\(\text{AP:} \ \dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}, \dfrac{1}{d} \ \ \text{where common difference}=k\ \ (k>0)\)

\(\text{Show} \ \ b+c<a+d\)

\(\dfrac{1}{a}<\dfrac{1}{a}+k\ \ (k>0)\)

\(\dfrac{1}{a}<\dfrac{1}{b}\)

\(\text{Similarly for}\ \ \dfrac{1}{b}<\dfrac{1}{c}\  \ldots\  \text{such that}\)

\(\dfrac{1}{a}<\dfrac{1}{b}<\dfrac{1}{c}<\dfrac{1}{d} \ \ \Rightarrow\ \ a>b>c>d\ \ldots\ (1)\)
 

\(\dfrac{1}{b}=\dfrac{1}{a}+k \ \ \Rightarrow \ \ k=\dfrac{1}{b}-\dfrac{1}{a}=\dfrac{a-b}{a b}\ \ \Rightarrow \ \ a-b=kab\)

\(\text {Similarly:}\)

\(\dfrac{1}{d}=\dfrac{1}{c}+k \ \ \Rightarrow \ \ k=\dfrac{c-d}{cd}\ \ \Rightarrow\ \ c-d=kcd\)
 

\(\text{Using \(\ a>b>c>d\ \)  (see (1) above):}\)

\(a-b>c-d\)

\(b+c<a+d\)

Proof, EXT2 P1 2024 HSC 13d

It is known that for all positive real numbers \(x, y\)

\(x+y \geq 2 \sqrt{x y} .\)     (Do NOT prove this.)

Show that if \(a, b, c\) are positive real numbers with  \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)  then  \(a \sqrt{b c}+b \sqrt{a c}+c \sqrt{a b} \leq a b c\).   (3 marks)

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\(\text{If }\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1, \ \ \ a, b, c \in \mathbb{R} ^{+}\)

\(\text {Prove} \ \ a \sqrt{b c}+b \sqrt{a c}+c \sqrt{a b} \leqslant a b c\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) \(=1\)  
\(\dfrac{b c+a c+a b}{a b c}\) \(=1\)  
\(ab+bc+ac\) \(=abc\ …\ (1)\)  

 
\(x+y \geqslant 2 \sqrt{x y}, \quad x, y \in \mathbb{R}^{+} \text{(given)}\)

\(a+b \geqslant 2 \sqrt{a b} \ \Rightarrow \ \sqrt{a b} \leqslant \dfrac{1}{2}(a+b)\)

\(\text{Similarly,}\)

\(\sqrt{bc} \leqslant \dfrac{1}{2}(b+c) \ \text{and}\ \ \sqrt{a c} \leqslant \dfrac{1}{2}(a+c)\)

  \(a \sqrt{b c}+b \sqrt{a c}+c \sqrt{a b}\) \(\leqslant a \times \dfrac{1}{2}(b+c)+b \times \dfrac{1}{2}(a+c)+c \times \dfrac{1}{2}(a+b)\)
    \(\leqslant \dfrac{1}{2} a b+\dfrac{1}{2} a c+\dfrac{1}{2} a b+\dfrac{1}{2} b c+\dfrac{1}{2} a c+\dfrac{1}{2} b c\)
    \(\leqslant a b+b c+a c\)
    \(\leqslant a b c\ \  \text{(see (1) above)}\)

Show Worked Solution

\(\text{If }\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1, \ \ \ a, b, c \in \mathbb{R} ^{+}\)

\(\text {Prove} \ \ a \sqrt{b c}+b \sqrt{a c}+c \sqrt{a b} \leqslant a b c\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) \(=1\)  
\(\dfrac{b c+a c+a b}{a b c}\) \(=1\)  
\(ab+bc+ac\) \(=abc\ …\ (1)\)  

 
\(x+y \geqslant 2 \sqrt{x y}, \quad x, y \in \mathbb{R}^{+} \text{(given)}\)

\(a+b \geqslant 2 \sqrt{a b} \ \Rightarrow \ \sqrt{a b} \leqslant \dfrac{1}{2}(a+b)\)

\(\text{Similarly,}\)

\(\sqrt{bc} \leqslant \dfrac{1}{2}(b+c) \ \text{and}\ \ \sqrt{a c} \leqslant \dfrac{1}{2}(a+c)\)

  \(a \sqrt{b c}+b \sqrt{a c}+c \sqrt{a b}\) \(\leqslant a \times \dfrac{1}{2}(b+c)+b \times \dfrac{1}{2}(a+c)+c \times \dfrac{1}{2}(a+b)\)
    \(\leqslant \dfrac{1}{2} a b+\dfrac{1}{2} a c+\dfrac{1}{2} a b+\dfrac{1}{2} b c+\dfrac{1}{2} a c+\dfrac{1}{2} b c\)
    \(\leqslant a b+b c+a c\)
    \(\leqslant a b c\ \  \text{(see (1) above)}\)

Proof, EXT2 P1 2023 HSC 16b

  1. Prove that  \(x>\ln x\), for  \(x>0\).  (2 marks)

  2. Using part (i), or otherwise, prove that for all positive integers \(n\),

\(  e^{n^2+n}>(n !)^2 .\)  (3 marks)

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i.   \(\text{See Worked Solutions}\)

ii.  \(\text{See Worked Solutions}\)

Show Worked Solution

i.    \(\text{Prove}\ \ x > \ln x\ \ \text{for} \ \ x>0: \)

\(\Rightarrow \ \text{Show}\ \ f(x) = x-\ln x > 0 \)

\(\text{SP’s occur when}\ \ f^{′}(x) = 1-\dfrac{1}{x} = 0\)

\(\text{SP at}\ (1,1) \)

\(f^{″} = x^{-2}>0,\ \ \forall x>0 \)

\(\text{SP at (1, 1) is a global minimum for}\ x>0 \)

\(\Rightarrow f(x) \geq 1 > 0 \)

\(\therefore x > \ln x\ \ \text{for} \ \ x>0 \)
 

♦ Mean mark (i) 39%.

ii.    \(x > \ln x\ \ \text{for} \ \ x>0 \ \ \Rightarrow \ \ e^x > x\ \ \text{(by definition)} \)

\(\text{Choose any positive integer}\ n: \)

\(e^n\) \(>n \)  
\(e^{n-1}\) \(>n-1 \)  
\(\ \ \vdots \)    
\(e^2\) \(>2\)  
\(e^1\) \(>1\)  

 
\(\text{Multiply each side of the equations above:}\)

\(e^n \times e^{n-1} \times \cdots \times e^{1} \) \(>n(n-1)(n-2) \cdots (2)(1) \)  
\(e^{n+(n-1)+(n-2)+ \cdots + 2+1}\) \(>n!\)  
\(e^{\frac{n(n+1)}{2}} \) \(>n!\ \ (\text{using AP formula}\ \ S_n=\frac{n}{2}(a+l) ) \)  
\(e^{\frac{n(n+1)}{2} \times 2}\) \(>(n!)^2\)  
\(e^{n^2+n}\) \(>(n!)^2\ \ …\ \text{as required}\)  
♦♦ Mean mark (ii) 28%.

Proof, EXT2 P1 2022 HSC 16c

It is given that for positive numbers `x_(1),x_(2),x_(3),dots,x_(n)` with arithmetic mean `A`,
 

               `(x_(1)xxx_(2)xxx_(3)xx cdots xxx_(n))/(A^(n)) <= 1`    (Do NOT prove this.)

Suppose a rectangular prism has dimensions  `a,b,c`  and surface area `S`.

  1. Show that  `abc <= ((S)/(6))^((3)/(2))`.  (2 marks)

  2. Using part (i), show that when the rectangular prism with surface area `S` is a cube, it has maximum volume.  (2 marks)

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  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
Show Worked Solution

i.    `text{Show}\ \ abc <= ((S)/(6))^((3)/(2))`

`S=2(ab+bc+ac)`

`A=(ab+bc+ac)/3`

`text{Using given relationship, where}\ x_1=ab, x_2=bc, …`


♦♦♦ Mean mark (i) 26%.
`(ab xx bc xx ca)/((ab+bc+ac)/3)^3` `<=1`  
`(ab xx bc xx ca)` `<=((ab+bc+ac)/3)^3`  
`(abc)^2` `<=((2(ab+bc+ac))/6)^3`  
`abc` `<=(S/6)^(3/2)`  

 

ii.   `text{If prism is a cube}\ \ a=b=c`

`=> V=a^3, \ \ S=6a^2`

`(S/6)^(3/2)=((6a^2)/6)^(3/2)=a^3`
 

`text{In the case of a cube:}`

`V=(S/6)^(3/2)`


♦♦♦ Mean mark (ii) 26%.

`text{Also,}\ \ V<=(S/6)^(3/2)\ \ \ text{(using part (i))}`

`:.\ text{For rectangular prisms with a given surface area, a cube}`

`text{has the maximum volume.}`

Proof, EXT2 P1 2021 HSC 15a

For all non-negative real numbers `x` and `y, \ sqrt(xy) <= (x + y)/2`.  (Do NOT prove this.)

  1. Using this fact, show that for all non-negative real numbers `a`, `b` and `c`,
  2.     `sqrt(abc) <= (a^2 + b^2 + 2c)/4`.  (2 marks)

  3. Using part (i), or otherwise, show that for all non-negative real numbers `a`, `b` and `c`,  
  4.     `sqrt(abc) <= (a^2 + b^2 + c^2 + a + b + c)/6`.  (2 marks)

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  1. `text(See Worked Solution)`
  2. `text(See Worked Solution)`
Show Worked Solution

i.   `text(Show)\ \ sqrt(abc) <= (a^2 + b^2 + 2c)/4`

♦ Mean mark part (i) 50%.

`sqrt(abc) = sqrt((ab)c) <= (ab + c)/2\ …\ (1)`
 

`text(Let)\ \ x = a^2\ \ text(and)\ \ y = b^2:`

`sqrt(xy) = sqrt(a^2b^2) ` `<= (a^2 + b^2)/2`
`ab` `<= (a^2 + b^2)/2\ …\ (2)`

 
`text{Substitute (2) into (1):}`

`sqrt(abc) <= ((a^2 + b^2)/2 + c)/2`

`sqrt(abc) <= (a^2 + b^2 + 2c)/4`

 

ii.   `text(Show)\ sqrt(abc) <= (a^2 + b^2 + c^2 + a + b + c)/6`

♦♦♦ Mean mark part (ii) 26%.

`text{Similarly (to part a):}`

`text(If)\ \ x = b^2\ \ text(and)\ \ y = c^2`

  `sqrt(abc) <= (b^2 + c^2 + 2a)/4`

`text(If)\ \ x = c^2\ \ text(and)\ \ y = a^2`

  `sqrt(abc) <= (c^2 + a^2 + 2b)/4`

`:.3sqrt(abc)` `<= (a^2 + b^2 + 2c + b^2 + c^2 + 2a + c^2 + a^2 + 2b)/4`
`3sqrt(abc)` `<= (2(a^2 + b^2 + c^2 + a + b + c))/4`
`sqrt(abc)` `<= (a^2 + b^2 + c^2 + a + b + c)/6`

Proof, EXT2 P1 SM-Bank 6

If  `x,  y,  z  ∈ R`  and  `x  ≠ y ≠ z`, then   

  1.  Prove  `x^2 + y^2 + z^2 > yz + zx + xy`   (2 marks)

  2.  If  `x + y + z = 1`, show  `yz+zx+xy<1/3`   (2 marks)

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  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
Show Worked Solution

i.   `x^2 + y^2 + z^2 – yz – zx – xy > 0`

`text(Multiply) × 2`

`2x^2 + 2y^2 + 2z^2 – 2yz – 2zx – 2xy > 0`

`(x^2 – 2xy + y^2) + (y^2 – 2yz + z^2) + (z^2 -2xz + x^2) > 0`

`(x – y)^2 + (y – z)^2 + (z – x)^2 > 0`

`text(Square of any rational number) > 0`

`:.\ text(Statement is true.)`

 

ii.    `(x + y + z)^2` `= x^2 + xy + xz + yx + y^2 + yz + zx + zy + z^2`
  `1` `= x^2 + y^2 + z^2 + 2xy + 2yz + 2xz`

 
`text{Consider statement in part (i):}`

`x^2 + y^2 + z^2 – yz – zx – xy > 0`

`=> (x + y + z)^2 – (x^2 + y^2 + z^2 – yz – zx – xy)<1`

`3xy + 3yz + 3zx` `< 1`
`:. \ yz + zx + xy` `< (1)/(3)`

Proof, EXT2 P1 2006 HSC 8a

Suppose  `0 <= t <= 1/sqrt 2.`

  1. Show that  `0 <= (2t^2)/(1 - t^2) <= 4t^2.`   (2 marks)

  2. Hence show that  `0 <= 1/(1 + t) + 1/(1 - t) - 2 <= 4t^2.`   (1 mark)

  3. By integrating the expressions in the inequality in part (ii) with respect to  `t`  from  `t = 0`  to  `t = x\ \  text{(where}\ \ 0 <= x <= 1/sqrt2\ \ text{)}`, show that
     
        `0 <= log_e ((1 + x)/(1 - x)) - 2x <= (4x^3)/3.`  (2 marks)

  4. Hence show that for  `0 <= x <= 1/sqrt 2`
     
        `1 <= ((1 + x)/(1 - x)) e^(-2x) <= e^((4x^3)/3).`  (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
i.    `0` `<= t <= 1/sqrt 2`
  `0` `<= t^2 <= 1/2`
  `0` `>= -t^2 >= -1/2`
  `1` `>= 1 – t^2 >= 1/2`
  `1` `<= 1/(1 – t^2) <= 2`
  `2t^2` `<= (2t^2)/(1 – t^2) <= 4t^2,\ \ \ \ (2t^2>0)`
  `:. 0` `<= (2t^2)/(1 – t^2) <= 4t^2`

 

ii.   `1/(1 + t) + 1/(1 – t) – 2`

`=((1-t)+(1+t)-2(1-t^2))/(1-t^2)`

`=(2t^2)/(1-t^2)`

 

`text{Substituting into part (i)}`

`:. 0 <= 1/(1 + t) + 1/(1 – t) – 2 <= 4t^2`

 

iii.   `int_0^x 0\ dt` `<= int_0^x (1/(1 + t) + 1/(1 – t) – 2)\ dt <= int_0^x 4t^2\ dt`
  `0` `<= [log_e (1 + t) – log_e (1 – t) – 2t]_0^x <= [(4t^3)/3]_0^x`
  `0` `<= [log_e (1 + x) – log_e (1 – x) – 2x]<= [(4x^3)/3]`
  `0` `<= log_e ((1 + x)/(1 – x)) – 2x <= (4x^3)/3`

 

iv.   `text(S)text(ince)\ \ e^a>e^b\ \ text(when)\ \ a>b` 

`e^0` `<= e^([ln ((1 + x)/(1 – x)) – 2x]) <= e^((4x^3)/3)`
`1` `<= e^(ln ((1 + x)/(1 – x))) xx e^(-2x) <= e^((4x^3)/3)`
`1` `<= ((1 + x)/(1 – x)) e^(-2x) <= e^((4x^3)/3)`

 

Proof, EXT2 P1 2011 HSC 5b

If  `p, q`  and  `r`  are positive real numbers and  `p + q >= r`, prove that

`p/(1 + p) + q/(1 + q) - r/(1 + r) >= 0.`  (3 marks)

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`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

`text(If)\ \ p + q >= r,\ \ text(then)\ \ p + q – r >= 0`

♦ Mean mark 41%.
`text(LHS)` `= p/(1 + p) + q/(1 + q) – r/(1 + r)`
  `= (p(1 + q)(1 + r) + q(1 + p)(1 + r) – r(1 + p)(1 + q))/((1 + p)(1 + q)(1 + r))`
  `= (p(1 + q + r + qr) + q(1 + p + r + pr) – r(1 + p + q + pq))/((1 + p)(1 + q)(1 + r))`
  `= (p + pq + pr + pqr + q + pq + qr + pqr – r – pr – qr – pqr)/((1 + p)(1 + q)(1 + r))`
  `= ((p + q – r) + pq(2 + r))/((1 + p)(1 + q)(1 + r))`
  `>=(pq(2 + r))/((1 + p)(1 + q)(1 + r))\ \ \ \ text{(S}text{ince}\ \ p + q – r >= 0 text{)}`
  `>= 0\ \ \ \ \ \ \ text{(S}text{ince}\ \ p > 0,\ q > 0,\ r > 0text{)}`

Proof, EXT2 P1 2014 HSC 15a

Three positive real numbers `a`, `b` and `c` are such that  `a + b + c = 1`  and  `a ≤ b ≤ c`.

By considering the expansion of  `(a + b + c)^2`, or otherwise, show that 

`qquad 5a^2 + 3b^2 +c^2 ≤ 1`.  (2 marks)

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`text{Proof (See Worked Solutions)}`

 

Show Worked Solution

`a + b+ c = 1,\ \ a ≤ b ≤ c`

♦♦ Mean mark 13%.

`(a + b+ c)^2 = 1`

`a^2 + b^2 + c^2 + 2ab+ 2bc + 2ac = 1`

`text(S)text(ince)\ a ≤ b\ \ =>a^2 ≤ ab`

`text(S)text(ince)\ b ≤ c\ \ =>b^2 ≤ bc`

`text(S)text(ince)\ a ≤ c\ \ =>a^2 ≤ ac`
 

`=> a^2 + b^2 + c^2 + 2a^2 + 2b^2 + 2a^2` `≤ 1`
`:.5a^2 + 3b^2 + c^2` `≤ 1\ \ \ text(… as required)`