2021/094) Let a and b be positive real numbers such that $a+b=1$. Prove that $a^a+b^b <=1$

We are given 

$1= a+ b = a^{a+b} + b^{a+b}$

So $1- (a^ab^b + a^b b^a)$

$=  a^{a+b} + b^{a+b} - (a^ab^b + a^b b^a)$

$= a^a(a^b-b^b) + b^a(b^b-a^b) = (a^a - b^a)(a^b - b^b)$

For a > b both the terms are non -ve so we have and if b > a then both terms are -ve and hence above is positive

$1- (a^ab^b + a^b b^a) >=0$ and hence the result

2021/093) Given $2^x = 3^y = 6^{-z}$ evaluate $\frac{1}{x} +\frac{1}{y} + \frac{1}{z}$

we are given $2^x = 3^y = 6^{-z}$

so $2 = 6^{\frac{-z}{x}}$

and $3 = 6^\frac{-z}{y}$

so $2 * 3 = 6^{(-\frac{z}{x} - \frac{z}{y})}$

or $6^1 = 6^{(-\frac{-z}{x} - \frac{z}{y})}$

or $1 = -\frac{z}{x} - \frac{z}{y}$

or  $\frac{z}{x} +\frac{z}{y} + 1 = 0 $

or $\frac{1}{x} +\frac{1}{y} + \frac{1}{z}= 0$

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 $x^2+y^2+z^2 \ge 3$