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Proof, EXT2 P1 2025 HSC 13c

  1. For positive real numbers \(a\) and \(b\), prove that  \(\dfrac{a+b}{2} \geq \sqrt{a b}\).   (2 marks)

  2. Hence, or otherwise, show that  \(\dfrac{2 n+1}{2 n+2}<\dfrac{\sqrt{2 n+1}}{\sqrt{2 n+3}}\)  for any integer \(n \geq 0\).   (2 marks)

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i.    \(\text {Using}\ \ (\sqrt{a}-\sqrt{b})^2 \geqslant 0:\)

\(a-2 \sqrt{a b}+b\) \(\geqslant 0\)
\(a+b\) \(\geqslant 2 \sqrt{a b}\)
\(\dfrac{a+b}{2}\) \(\geqslant \sqrt{a b}\)

 
ii.    \(\text{Show}\ \ \dfrac{2 n+1}{2 n+2}<\dfrac{\sqrt{2 n+1}}{\sqrt{2 n+3}}\)

\(\text{Using part (i):}\)

\(\text{Let} \ \ a=2 n+1, b=2 n+3\)

\(\dfrac{2 n+1+2 n+3}{2}\) \(\geqslant \sqrt{2 n+1} \sqrt{2 n+3}\)
\(2 n+2\) \(\geqslant \sqrt{2 n+1} \sqrt{2 n+3}\)
\(\dfrac{1}{2 n+2}\) \(\leqslant \dfrac{1}{\sqrt{2 n+1} \sqrt{2 n+3}}\)
\(\dfrac{2 n+1}{2 n+2}\) \(\leqslant \dfrac{\sqrt{2 n+1}}{\sqrt{2 n+3}}\)

 

\(\text{Consider part (i):}\ (\sqrt{a}-\sqrt{b})^2 \geqslant 0\)

\(\Rightarrow \ \text{Equality only holds when} \ \ a=b\)

\(\text{Since}\ \ a=2 n+1 \neq b=2 n+3\)

\(\dfrac{2 n+1}{2 n+2}<\dfrac{\sqrt{2 n+1}}{\sqrt{2 n+3}}\)

Proof, EXT2 P1 2023 HSC 12b

Prove that for all real numbers \(x\) and \(y\), where \(x^2+y^2 \neq 0\),

\(\dfrac{(x+y)^2}{x^2+y^2} \leq 2 \text {. }\)  (2 marks)

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

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\((x-y)^2 \) \(\geq 0 \)  
\(x^2-2xy+y^2 \) \(\geq 0\)  
\(x^2+y^2\) \(\geq 2xy \)  
\( \dfrac{2xy}{x^2+y^2}\) \( \leq 1\ \text{… (1)}\)  

 

\(\dfrac{(x+y)^2}{x^2+y^2}\) \(=\dfrac{x^2+2xy+y^2}{x^2+y^2}\)  
  \(=\dfrac{x^2+y^2}{x^2+y^2}+\underbrace{\dfrac{2xy}{x^2+y^2}}_{\text{see (1) above}} \)  
  \(\leq 1+1\)  
  \(\leq 2\)  

Proof, EXT2 P1 2022 HSC 12a

For real numbers  `a,b >= 0`  prove that  `(a+b)/(2) >= sqrt(ab)`.  (2 marks)

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`text{Proof (See Worked Solutions)}`

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`text{S}text{ince}\ \ (sqrta-sqrtb)^2>=0:`

`a-2sqrt(ab)+b` `>=0`  
`a+b` `>=2sqrt(ab)`  
`:.(a+b)/2` `>=sqrt(ab)\ \ text{… as required}`  

Mean mark 93%.

Proof, EXT2 P1 2020 HSC 13c

  1. By considering the right-angled triangle below, or otherwise, prove that
     
         `frac{a + b}{2} ≥ sqrt(ab)`, where  `a > b ≥ 0`.  (2 marks)
     
           

  2. Prove that  `p^2 + 4q^2 ≥ 4 pq`.  (1 marks)

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

i.   `text{Strategy 1}`

`text{Using Pythagoras:}`

`x` `= sqrt{(a + b)^2-(a-b)^2}`
  `= sqrt(4ab)`
  `= 2 sqrt(ab)`

  
`a + b \ text{is a hypotenuse}`

`a + b` `≥ x`
`a + b` `≥ 2sqrt(ab)`
`frac{a + b}{2}` `≥ sqrt(ab)`

 

`text{Strategy 2}`

`(sqrta-sqrtb)^2` `≥ 0`
`a-2 sqrt(a) sqrt(b) + b` `≥ 0`
`a + b` `≥ 2 sqrt(ab)`
`frac{a + b}{2}` `≥ sqrt(ab)`

 

ii.    `text{Let} \ \ a = p  ,  b = 2 q`

`text(Using part i:)`

`frac{p + 2q}{2}` `≥ sqrt(2 pq)`
`p + 2q` `≥ 2 sqrt(2 pq)`
`p^2 + 4pq + 4 q^2` `≥ 8 pq`
`p^2 + 4 q^2` `≥ 4 pq`

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 2018 HSC 9 MC

It is given that  `a`, `b` are real and  `p`, `q` are purely imaginary.

Which pair of inequalities must always be true?

  1. `a^2p^2 + b^2q^2 <= 2abpq,qquada^2b^2 + p^2q^2 <= 2abpq`
  2. `a^2p^2 + b^2q^2 <= 2abpq,qquada^2b^2 + p^2q^2 >= 2abpq`
  3. `a^2p^2 + b^2q^2 >= 2abpq,qquada^2b^2 + p^2q^2 <= 2abpq`
  4. `a^2p^2 + b^2q^2 >= 2abpq,qquada^2b^2 + p^2q^2 >= 2abpq`
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`B`

Show Worked Solution

`a, b ->\ text(real)qquad\ p, q ->\ text(purely imaginary)`

`=> ab, pq, (ab – pq)\ text(are real)`

`=> ap, bq, (ap – bq)\ text(are purely imaginary.)`
 

`(ab – pq)^2` `>= 0`
`a^2b^2 + p^2q^2` `>= 2abpq\ \ (text(Eliminate A and C))`

 

`(ap – bq)^2` `<= 0`
`a^2p^2 + b^2q^2` `<= 2abpq`

 
`=>B`

Proof, EXT2 P1 2016 HSC 14c

Show that  `x sqrt x + 1 >= x + sqrt x,`  for  `x >= 0.`  (3 marks)

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`text(See Worked Solutions)`

Show Worked Solution

`text(Show)\ \ xsqrtx + 1 >= x + sqrtx,quadtext(for)\ x >= 0`

♦ Mean mark 39%.
`(text(LHS))^2` `= x^3 + 2xsqrtx + 1`
  `= 2xsqrtx + (x + 1)(x^2 – x + 1)`

 

`(x – 1)^2` `>= 0`
`x^2 – 2x + 1` `>= 0`
`:. x^2 + 1` `>= 2x`

 

`:.\ (text(LHS))^2` `>= 2xsqrtx + (x + 1)(2x – x)`
  `>= 2xsqrtx + x(x + 1)`
  `>= x^2 +2xsqrtx+x`
  `>= (x + sqrtx)^2`
  `>= (text(RHS))^2`

 

`:.\ text(LHS ≥ RHS for)\ \ x >= 0`

Proof, EXT2 P1 2015 HSC 15c

For positive real numbers `x` and `y`, `sqrt (xy) <= (x + y)/2`.     (Do NOT prove this.)

  1. Prove  `sqrt (xy) <= sqrt ((x^2 + y^2)/2)`,  for positive real numbers  `x`  and  `y.`  (1 mark)

  2. Prove  `root4(abcd) <= sqrt ((a^2 + b^2 + c^2 + d^2)/4)`,  for positive real numbers  `a, b, c`  and  `d.`  (2 marks)

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

i.    `text(S)text(ince)\ \ (x – y)^2 = x^2 + y^2 – 2xy`

`and \ (x – y)^2 >= 0`

`0` `≤x^2 + y^2 – 2xy`
`2xy` `≤x^2 + y^2`
`:.sqrt (xy)` `≤sqrt ((x^2 + y^2)/2)`

 

ii.   `sqrt(ab) <= sqrt((a^2 + b^2)/2),\ \ \ \ text{(part (i))}`

♦♦ Mean mark 29%. 

`sqrt(cd) <= sqrt((c^2 + d^2)/2),\ \ \ \ text{(part (i))}`

`sqrt(ab) sqrt(cd)` `<=sqrt((a^2 + b^2)/2)*sqrt((c^2 + d^2)/2)`
  `<=sqrt(((a^2 + b^2)/2) * ((c^2 + d^2)/2))`
 
`sqrt (xy) <= (x + y)/2\ \ \ \ text{(given):}`
 `sqrt(ab) sqrt(cd)`  `<=1/2((a^2 + b^2+c^2+d^2)/2)`
`sqrt(abcd)` `<=(a^2 + b^2+c^2+d^2)/4`
`:.root4(abcd)`  `<=sqrt((a^2 + b^2+c^2+d^2)/4)`

Proof, EXT2 P1 2012 HSC 15a

  1. Prove that  `sqrt(ab) ≤ (a + b)/2`, where  `a ≥ 0`  and  `b ≥ 0`.   (1 mark)

  2. If  `1 ≤ x ≤ y`,  show that  `x(y − x + 1) ≥ y`.   (2 marks)

  3. Let  `n`  and  `j`  be positive integers with  `1 ≤ j ≤ n`.
     
    Prove that  `sqrtn ≤ sqrt(j(n − j + 1)) ≤ (n + 1)/2.`  (2 marks)

  4. For integers  `n ≥ 1`, prove that
     
        `(sqrtn)^n ≤ n! ≤ ((n + 1)/2)^n`.   (1 mark)

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  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
  3. `text{Proof (See Worked Solutions)}`
  4. `text{Proof (See Worked Solutions)}`
Show Worked Solution
i. `(sqrta − sqrtb)^2` `≥ 0`
  `a − 2sqrt(ab) + b` `≥ 0`
  `a + b` `≥ 2sqrt(ab)`
  `sqrt(ab)` `≤ (a + b)/2`

 

ii.  `text(Solution 1)`

♦♦ Mean mark part (ii) 28%.

`text(S)text(ince)\ \ 1 ≤ x ≤ y`

`y-x` `>=0`
`y(x-1)-x(x-1)` `>=0,\ \ \ \ (x-1>=0)`
`xy-x^2+x-y` `>=0`
`:.xy-x^2+x` `>=y`

 

`text(Solution 2)`

`x( y − x + 1)` `= xy − x^2 + x`
  `= -y + xy − x^2 + x + y`
  `= y(x − 1) − x(x − 1)+ y`
  `= (x − 1)( y − x) + y`

 

`text(S)text(ince)\ \ x − 1 ≥ 0\ \ text(and)\ \  y − x ≥ 0`

`=>(x − 1)( y − x) + y` `>=y`
`:.x(y − x + 1)` `>=y`

 

iii.  `text(Let)\ c = n − j + 1, \ c > 0\ text(as)\ n ≥ j.`

♦ Mean mark part (iii) 39%.

 

`=> sqrt(j(n − j + 1))\ \ text(can be written as)\ \ sqrt(jc).`

`sqrt(jc)` `≤ (j + c)/2\ \ \ \ text{(part (i))}`
`sqrt(j(n − j + 1))` `≤ (j + n-j+1)/2`
`=>sqrt(j(n − j + 1))` `≤ (n+1)/2`
`j(n − j +1)` `≥ n\ \ \ \ text{(part (ii))}`
`=>sqrt(j(n − j + 1))` `≥ sqrt n`

 

`:.sqrtn ≤ sqrt(j(n − j + 1)) ≤ (n + 1)/2`

 

iv. `text{From (iii)}\ n ≤ j(n− j +1) ≤ (n + 1)/2`

♦♦♦ Mean mark part (iv) just 2%!

`text(Let)\ j\ text(take on the values from 1 to)\ n.`

`j = 1:` `sqrtn ≤ sqrt(1(n)) ≤ (n + 1)/2`
`j = 2:` `sqrtn ≤ sqrt(2(n−1)) ≤ (n + 1)/2`
  `vdots`
`j = n:` `sqrtn ≤ sqrt(n(1)) ≤ (n + 1)/2`

 

`text{Multiply the corresponding parts of each line}`

`(sqrtn)^n ≤ sqrt(n! xx n!) ≤ ((n + 1)/2)^n`

`(sqrtn)^n ≤ n! ≤ ((n + 1)/2)^n`