# Selected Problems on Inequality - Part I

Published by Ganit Charcha | Category - Math Articles | 2014-11-01 03:21:01

Problem 1: If $a$, $b$ and $c$ are three positivie numbers then, $$\frac{b^2 + c^2}{b + c} + \frac{c^2 + a^2}{c + a} + \frac{a^2 + b^2}{a + b} \geq a + b + c$$
Solution:
\begin{align}
\frac{b^2 + c^2}{b + c} + \frac{c^2 + a^2}{c + a} + \frac{a^2 + b^2}{a + b} - (a + b + c) = (\frac{b^2 + c^2}{b + c} - \frac{b + c}{2}) + \\(\frac{c^2 + a^2}{c + a} - \frac{c + c}{2}) + (\frac{a^2 + b^2}{a + b} - \frac{a + b}{2})
\nonumber
\end{align}
\begin{align}
\hspace{1.2in} = \frac{(b-c)^2}{2(b+c)} + \frac{(c-a)^2}{2(c+a)} + \frac{(a-b)^2}{2(a+b)} \geq 0
\end{align}
since a, b and c are positive numbers.

Problem 2: If $a$, $b$ and $c$ are the sides of a triangle and $x + y + z = 0$, then show that for $x \neq 0$, $y \neq 0$ and $z \neq 0$, $a^{2}yz + b^{2}zx + c^{2}xy \leq 0$
Solution:
Since $x + y + z = 0$ and $x \neq 0$, $y \neq 0$ and $z \neq 0$, so without loss of generality we can assume that $x$ and $y$ are positive and $z$ is negative.
Therefore, $z = -(x + y)$.
\begin{align}
a^{2}yz + b^{2}zx + c^{2}xy = -(a^{2}y + b^{2}x)(x + y) + c^{2}xy
\nonumber\\
= -a^{2}xy - b^{2}x^{2} - a^{2}y^{2} - b^{2}xy + c^{2}xy
\nonumber\\
= -xy[a^{2} + 2ab + b^{2} - c^{2}] + 2abxy - b^{2}x^{2} - a^{2}y^{2}
\nonumber\\
= -xy[(a + b)^{2} - c^{2}] - (ay - bx)^{2}
\end{align}
$x$ and $y$ are considered as positive and since $a$, $b$ and $c$ are the sides of a triangle, so $a + b > c$ and so $xy[(a + b)^{2} - c^{2}] > 0$ and being
a square $(ay - bx)^{2} \geq 0$ and hence from equation (6) it follows that $a^{2}yz + b^{2}zx + c^{2}xy \leq 0$.

Problem 3: If $a$, $b$ and $x$, $y$ are all positive numbers, then show that $\frac{(a + b)xy}{ay + bx} \leq \frac{ax + by}{a + b}$
Solution:
\begin{align}
\frac{(a + b)xy}{ay + bx} - \frac{ax + by}{a + b} = \frac{ab(2xy - x^{2} - y^{2})}{(a + b)(ay + bx)} = - \frac{ab(x - y)^{2}}{(a + b)(ay + bx)} \leq 0
\end{align}
since $a$, $b$, $x$ and $y$ are all positive numbers.

Problem 4: Prove that, $\frac{1}{2\sqrt{n+1}} < \frac{1}{2}.\frac{3}{4}.\frac{5}{6}\ldots\frac{2n - 1}{2n} < \frac{1}{\sqrt{2n+1}}$
Solution:
Let $u_{n} =\frac{1}{2}.\frac{3}{4}.\frac{5}{6}\ldots\frac{2n - 1}{2n}$.
Note that, $\frac{1}{2} < \frac{2}{3}, \frac{3}{4} < \frac{4}{5}, \frac{5}{6} < \frac{6}{7}, \ldots, \frac{2n-1}{2n} < \frac{2n}{2n+1}$. Combining them, we get
\begin{align}
\frac{1}{2}.\frac{3}{4}.\frac{5}{6}\ldots\frac{2n - 1}{2n} < \frac{2}{3}.\frac{4}{5}.\frac{6}{7}\ldots\frac{2n}{2n + 1}
=> u_{n} < \frac{2}{3}.\frac{4}{5}.\frac{6}{7}\ldots\frac{2n}{2n + 1}
\nonumber\\
u_{n}^{2} = u_{n}.u_{n} < (\frac{1}{2}.\frac{3}{4}.\frac{5}{6}\ldots\frac{2n - 1}{2n}).( \frac{2}{3}.\frac{4}{5}.\frac{6}{7}\ldots\frac{2n}{2n + 1}) = \frac{1}{2n + 1} \hspace{1cm} [Using \hspace{0.4cm} (4)]\\
u_n < \frac{1}{\sqrt{2n + 1}} \hspace{2.5in}
\end{align}
Also, $(2n + 1).u_{n} = \frac{1}{2}.\frac{3}{4}.\frac{5}{6}\ldots\frac{2n - 1}{2n}.(2n + 1) = \frac{3}{2}.\frac{5}{4}.\frac{7}{6}\ldots\frac{2n + 1}{2n}$
Note that, $\frac{3}{2} > \frac{4}{3}, \frac{5}{4} > \frac{6}{5}, \frac{7}{6} > \frac{8}{7}, \ldots, \frac{2n + 1}{2n} > \frac{2n + 2}{2n+1}$. Combining them, we get
\begin{align}
\hspace{-8in}\frac{3}{2}.\frac{5}{4}.\frac{7}{6}\ldots\frac{2n + 1}{2n} > \frac{4}{3}.\frac{6}{5}.\frac{8}{7}\ldots\frac{2n + 2}{2n + 1}\\
\hspace{-8in}=> (2n + 1).u_{n} > \frac{4}{3}.\frac{6}{5}.\frac{8}{7}\ldots\frac{2n + 2}{2n + 1}\\
\hspace{-0.2in}(2n + 1)^{2}.u_{n}^{2} =(2n + 1). u_{n}.(2n + 1).u_{n} > (\frac{4}{3}.\frac{6}{5}.\frac{8}{7}\ldots\frac{2n + 2}{2n + 1}).( \frac{3}{2}.\frac{5}{4}.\frac{7}{6}\ldots\frac{2n + 1}{2n}) = \frac{2n + 2}{2} \hspace{1cm} [Using \hspace{0.4cm} (6)]\\
\hspace{-4in}=> (2n + 1). u_{n} > \sqrt{n+1} \hspace{2.5in}\\
=> u_{n} > \frac{\sqrt{n+1}}{(2n + 1)} > \frac{\sqrt{n+1}}{(2n + 2)} \hspace{1cm} [Since \hspace{0.2cm} \frac{a}{b} > \frac{a}{b+1} \hspace{0.2cm} for \hspace{0.2cm} a > 0, b > 0] \hspace{1.5cm}\\
\hspace{-4.8in}=> u_{n} > \frac{1}{2\sqrt{n + 1}} \hspace{2.5in}
\end{align}
Combining, we have therefore $\frac{1}{2\sqrt{n+1}} < \frac{1}{2}.\frac{3}{4}.\frac{5}{6}\ldots\frac{2n - 1}{2n} < \frac{1}{\sqrt{2n+1}}$.

Problem 5: Prove that, $1!.3!.5!\ldots(2n-1)! > (n!)^{n}$
Solution:
If $n > r > 0$, then we have $2n - r > n$.
\begin{align}
n! = 1.2.3\ldots.r.(r+1).(r+2)\ldots.(n-1).n = r!.(r+1).(r+2)\ldots.(n-1).n
\end{align}
\begin{align}
(2n - r)! = 1.2.3\ldots.n.(n+1).(n+2)\ldots.(2n-r) = n!.(n+1).(n+2)\ldots.(2n-r)
\end{align}
\begin{align}
\frac{r!}{n!}.\frac{(2n - r)!}{n!} = \frac{(n+1).(n+2)\ldots.(2n-r)}{(r+1).(r+2)\ldots.(n-1).n} > 1
\nonumber\\
r!.(2n-r)! > (n!)^{2}
\end{align}
In case of $r > n$, in a similar way it can also be shown that $r!.(2n-r)! > (n!)^{2}$.
Putting $r = 1, 3, 5, \ldots, (2n - 1)$ successively and multiplying together we get $$(1!.3!.5!\ldots(2n-1)!)^{2} >( (n!)^{2})^{n}$$.
Taking the positive square root of both sides we get $$1!.3!.5!\ldots(2n-1)! > (n!)^{n}$$

Problem 6: Prove that, $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{n} < \sqrt{(2n - 1)}$.
Solution:
For $a > 0$ and $b > 0$, since arthmetic mean is always greater or equal to geometric mean, so we have $\frac{a + b}{2} \geq \sqrt{ab}$, equality holds when $a$ and $b$ are equal..
The following follows easily from this inequality.
\begin{align}
a + b + 2\sqrt{ab} \leq 2a + 2b
\nonumber\\
=> \frac{(\sqrt{a} + \sqrt{b})^2}{4} \leq \frac{2a + 2b}{4}
\nonumber\\
=> \frac{\sqrt{a} + \sqrt{b}}{2} \leq \sqrt{\frac{a + b}{2}}
\end{align}
Therefore, for $n \geq 2$, by choosing $a = 2n - 1 > 0$ and $b = 2n - 3 > 0$, using the above inequality we have the following
$\sqrt{2n - 1} + \sqrt{2n - 3} < 2\sqrt{\frac{(2n - 1) + (2n - 3)}{2}} = 2\sqrt{2n - 2} < 2n.$
The fact that $2\sqrt{2n - 2} < 2n$ follows from the fact that $(n- 1)^2 > 1 > -1$, since $n$ has been taken as greater or equal to $2$.
\begin{align}
\frac{1}{2n} < \frac{1}{\sqrt{2n - 1} + \sqrt{2n - 3}} = \frac{\sqrt{2n - 1} - \sqrt{2n - 3}}{(2n - 1) - (2n - 3)} = \frac{\sqrt{2n - 1} - \sqrt{2n - 3}}{2}
\nonumber\\
\hspace{-4.0in}=> \frac{1}{n} < \sqrt{2n - 1} - \sqrt{2n - 3}
\end{align}
Putting $n = 2, 3, \ldots, n$ and adding we get
\begin{align}
\frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n} = (\sqrt{3} - 1) + (\sqrt{5} - \sqrt{3}) + (\sqrt{7} - \sqrt{5}) + \ldots + (\sqrt{2n - 1} - \sqrt{2n - 3})
\nonumber\\
\hspace{-4.8in}=> 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n} < \sqrt{(2n - 1)}
\end{align}

Problem 7: Prove that, $(n!)^{2} > n^{n}$ if $n > 2$.
Solution:
\begin{align}
(n!)^{2} = 1.n.2.(n-1).3.(n-2).\ldots.r.(n - r + 1).\ldots.n.1
\end{align}
Now, let $n > r > 1$, then $n - r > 0$. Therefore, $(n-r).r > n - r$ and as such $r.(n - r + 1) > n$.
Putting $r = 2, 3,\ldots, r$, we get $2.(n-1) > n$, $3.(n-2) > n$, ...,$r.(n-r+1) > n$.
Hence, for $n > 2$ we have,
\begin{align}
(n!)^{2} = 1.n.2.(n-1).3.(n-2).\ldots.r.(n - r + 1).\ldots.n.1 > (n.n.n\ldots.n.\ldots.n) = n^{n}
\end{align}