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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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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\(\text{Proof (See Worked Solutions)}\)
\((x-y)^2 \) | \(\geq 0 \) | |
\(x^2-2xy+y^2 \) | \(\geq 0\) | |
\(x^2+y^2\) | \(\geq 2xy \) | |
\( \dfrac{2xy}{x^2+y^2}\) | \( \leq 1\ \text{… (1)}\) |
\(\dfrac{(x+y)^2}{x^2+y^2}\) | \(=\dfrac{x^2+2xy+y^2}{x^2+y^2}\) | |
\(=\dfrac{x^2+y^2}{x^2+y^2}+\underbrace{\dfrac{2xy}{x^2+y^2}}_{\text{see (1) above}} \) | ||
\(\leq 1+1\) | ||
\(\leq 2\) |
For real numbers `a,b >= 0` prove that `(a+b)/(2) >= sqrt(ab)`. (2 marks)
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`text{Proof (See Worked Solutions)}`
`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}` |
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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` |
If `x, y, z ∈ R` and `x ≠ y ≠ z`, then
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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)` |
It is given that `a`, `b` are real and `p`, `q` are purely imaginary.
Which pair of inequalities must always be true?
`B`
`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`
Show that `(r + s)/2 >= sqrt (rs)` for `r >= 0` and `s >= 0`. (1 mark)
`text(Proof)\ \ text{(See Worked Solutions)}`
`(sqrt r – sqrt s)^2` | `>= 0` |
`r – 2 sqrt r sqrt s + s` | `>= 0` |
`r + s` | `>= 2 sqrt r sqrt s` |
`(r + s)/2` | `>= sqrt (rs)` |
Show that `x sqrt x + 1 >= x + sqrt x,` for `x >= 0.` (3 marks)
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`text(See Worked Solutions)`
`text(Show)\ \ xsqrtx + 1 >= x + sqrtx,quadtext(for)\ x >= 0`
`(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`
For positive real numbers `x` and `y`, `sqrt (xy) <= (x + y)/2`. (Do NOT prove this.)
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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))}`
`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)` |
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i. | `(sqrta − sqrtb)^2` | `≥ 0` |
`a − 2sqrt(ab) + b` | `≥ 0` | |
`a + b` | `≥ 2sqrt(ab)` | |
`sqrt(ab)` | `≤ (a + b)/2` |
ii. `text(Solution 1)`
`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.`
`=> 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`
`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`