2021/092) The sum of two numbers is 15. What is the minimum sum of the resultant cubes of the two numbers?

Let the numbers be x and y

We have x+ y = 15

$(x+y)^3 = x^3 + y^3 + 3xy(x+y)$

or $15^3 = x^3 + y^3 + 3xy * 15$

or $x^3 + y^3 = 15^3 - 45xy$

this is mininum when xy is maximum

$x + y = 15$

we have $4xy = (x+y)^2 - (x-y)^2 = 15 - (-x-y)^2$

so xy is maximumum when x = y

or $x^3 + y^3$ is minumum when $x = y = 7.5$ and value is $2 * 7.5^3 = 843.75$

2021/091) if ab = cd prove that $a^2+b^2 + c^2 + d^2 $ is composite

 We have

ab = cd

or $\frac{a}{c}= \frac{d}{b} = \frac{m}{n}$ where m and n are in lowest terms or gcd(m,n) = 1

so an = cm and dn = bm

now $n^2(a^2+b^2+ c^2+d^2) $

$= (na)^2 + (nb)^2 + (nc)^2 + (nd)^2$

$= (cm)^2 + (bn)^2 +(nc)^2 + (nd)^2= (b^2+c^2)(m^2 + n^2)$

or $a^2+b^2+c^2+d^2 = \frac{b^2+c^2}{n^2} (m^2+n^2)$

as n is less than $b^2+c^2$ so it is product of 2 numbers > 1 so composite 

2021/090) If $p+q+r=0$ then what is $\frac{(p+q)^2}{3pq}+\frac{(q+r)^2}{3qr} +\frac{(r+p)^2}{3rp}$

We have p+q + r = 0

So $p^3 + q^3 + r^3 = 3pqr\cdots(1)$

And  p+q = -r

or $(p+q)^2 = r^2\cdots(2)$

Similary $(q+r)^2 = p^2\cdots(3)$

$(r+p)^2 = q^2\cdots(4)$

Hence $\frac{(p+q)^2}{3pq}+\frac{(q+r)^2}{3qr} +\frac{(r+p)^2}{3rp}$

$= \frac{r^2}{3pq}+\frac{p^2}{3qr} +\frac{q^2}{3rp}$ (from (2), (3), (4)

$= \frac{r^3+q^3+ p^3}{3pqr} = \frac{3pqr}{3pqr} = 1$  (using (1))

2021/089) Show that $2^n$ is not a factor of $3^n+1$ for n >1

We shall prove the same for 2 cases 

1) n is even

2) n is odd 

let us 1st prove for n even

case 1 :For n even say 2k $(>=2)$

$3^n + 1 =  3^{2k} + 1 = 9^k + 1 \equiv 2 \pmod 4$

so $3^n + 1$ is not divisible by 4 so cannot be divisible by $2^n$

    

case 2: for n odd  say 2k + 1 $( >=3)$

$3^n + 1 = 3^{2k+ 1} + 3 = 9^k.3 + 1 \equiv 4 \pmod 8$

so  so $3^n + 1$ is not divisible by 8 so cannot be divisible by $2^n$

2021/088) For real x,y z if x +y + z = 3 prove that $x^2+y^2 + z^2 >= 3$

We are given 

$x+y+z = 3 \cdots(1)$

now $(x-1)^2\ge 0$

or $x^2-2x + 1 \ge 0$

or $x^2 \ge 2x -1\cdots(2)$

similarly $y^2 \ge 2y-1\cdots(3)$

and $z^2\ge 2z -1\cdots(4)$

addding (2), (3), (4) we get $x^2+y^2 + z^2\ge 2(x+y+z) - 3$

or $x^2+y^2 + z^2\ge 2* 3- 3$ (from (1)

or $x^2+y^2+z^2 \ge 3$ 

2021/087)Given: $x>0,\, n\in\mathbb{N}$ Prove: $(1+x)\times\left(1+x^2 \right)\times\cdots\times\left(1+x^n \right)\geq\left(1+x^{\large{\frac{n+1}{2}}} \right)^n$

We have $x^a + x^b >= 2x^{\frac{a+b}{2}}$ by AM GM inequality

adding $1 + x^{a+b}$ on both sides we get

$(1+x^a)(1+x^b) >= 1 + 2x^{\frac{a+b}{2}} + x^{a+b}= ( 1+ x^{\frac{a+b}{2}})^2$

Putting $b = n+ 1 -a$ we get

$(1+x^a)(1+x^(n+1- a) >= (1+ x^{\frac{n+1}{2}})^2$

For n even taking a from 1 to $\frac{n}{2}$ we get n/2 expressions and multiplying them out we get the result

For n odd we have n-1 ( running a from 1 to $\frac{n-1}{2}$ we get $\frac{n-1}{2}$ terms and as middle term is $(1+x^{\frac{n+1}{2}})$ we get thr result

2021/086) Find the limit of: $lim_{x\to +\infty}\frac{ \lfloor x\rfloor}{x}$

We have

 $\frac{ \lfloor x\rfloor}{x} = \frac{ x- y}{x}$ where y is fractional part of x and hence $0 \le y \lt 1$

Hence 

$  \frac{ \lfloor x\rfloor}{x} =   1- \frac{y}{x} \ge 1 - \frac{1}{x}$

as x goes to infinitte RHS goes to 1 so the required value is between 1 and 1 so it is 1