Indian Statistical Institute ( ISI ) B.Math & B.Stat : Algebra

Let $a, b, c$ be real numbers greater than $1.$ Let $S$ denote the sum $S = log_{a}{bc} + log_{b}{ca} + log_{c}{ab}.$ Find the smallest possible value of $S.$ $S =log_{a}{\frac{abc}{a}} + log_{b}{\frac{bca}{b}} + log_{c}{\frac{abc}{c}} = log_{a}{abc} + log_{b}{abc} + log_{c}{abc}-3$ $= log_{a}{e} \times log_{e}{abc} + log_{b}{e} \times log_{e}{abc} + log_{c}{e} \times log_{e}{abc}-3$ $= log_{e}{abc} \bigg( \frac{1}{log_{e}{a}} +\frac{1}{log_{e}{b}}+\frac{1}{log_{e}{c}} \bigg) -3$ $= (log_{e}{a}+log_{e}{b}+log_{e}{c}) \bigg( \frac{1}{log_{e}{a}} +\frac{1}{log_{e}{b}}+\frac{1}{log_{e}{c}} \bigg) -3$ Now Using the inequality $A.M \times H.M \geq n^2$ for $n$ positive real numbers, we see that $(log_{e}{a}+log_{e}{b}+log_{e}{c}) \bigg( \frac{1}{log_{e}{a}} +\frac{1}{log_{e}{b}}+\frac{1}{log_{e}{c}} \bigg) \geq 3^2 = 9$ Thus $S \geq 9-3 = 6$. Note $log_{e}{a},log_{e}{b},log_{e}{c}$ are all positive since $a,b,c > 1$.

Indian Statistical Institute ( ISI ) B.Math & B.Stat : Algebra

Show that the polynomial $x^8 − x^7 + x^2 − x + 15$ has no real root. Let $f(x)=x^8 − x^7 + x^2 − x + 15$, we will show that $f(x) >0$ for all $x \in \mathbb{R}$. $f(x) = x^7(x-1)+x(x-1)+15 = (x-1)x(x^6+1)+15$ Now, $\mathbb{R} = (- \infty , 0] \cup (0,1] \cup (1, \infty)$. Note that $f(0)=f(1)=15$ When $x \in (1, \infty),$ $x,x-1 \quad and \quad x^6+1 > 0 \implies f(x) > 15 \quad \forall \quad x \in (1, \infty)$ When $x \in (- \infty , 0),$ $x \quad and \quad x-1 < 0 \quad thus \quad x(x-1) > 0$ since $x^6+1 > 0$ for any $x$, $f(x) > 15$ in this case too. When $x \in (0,1)$, $1 < x^6+1 < 2$ and $0 < x < 1$. Since both of them are positive $0 < x(x^6+1) < 2$. Further $-1 < x-1 < 0$, thus $x(x-1)(x^6+1) < 0$. Again $|x-1| < 1$ this implies $-2 < x(x-1)(x^6+1) < 0$. Thus $f(x) > 13 > 0$. Combining all the cases we see that $f(x) > 0 \quad \forall \quad x \in \mathbb{R}$ which shows $f(x)$ has no real root.

Indian Statistical Institute (ISI) B.Math & B.Stat : Combinatorics

For $k \geq 1$, find the value of $$\binom{n}{0}+ \binom{n+1}{1}+ \binom{n+2}{2}+ \dots + \binom{n+k}{k}$$ Using the identity $\binom{n}{r} = \binom{n}{n-r}$, $\binom{n}{0}+ \binom{n+1}{1}+ \binom{n+2}{2}+ \dots + \binom{n+k}{k}$ reduces to $$\binom{n}{n}+ \binom{n+1}{n}+ \binom{n+2}{n}+ \dots + \binom{n+k}{n}$$ = Coefficient of $x^n$ in $(1+x)^n$ + Coefficient of $x^n$ in $(1+x)^{n+1}$ + $\dots$ + Coefficient of $x^n$ in $(1+x)^{n+k}$ = Coefficient of $x^n$ in $(1+x)^n + (1+x)^{n+1} + \dots + (1+x)^{n+k}$ = Coefficient of $x^n$ in $(1+x)^n \frac{(1+x)^{k+1}-1}{1+x-1} = \frac{(1+x)^{n+k+1}-(1+x)^n}{x}$ = Coefficient of $x^{n+1}$ in $(1+x)^{n+k+1}$ $= \binom{n+k+1}{n+1}$