SEQUENCES AND SERIES - 4 (Inequalities based on A.M., G.M. and H.M.)
Properties of A.M, G.M & H.M
Let A, G, $\mathrm{H}$ be the arithmetic, geometric and harmonic means of two positive numbers a & (b).
Then, $A=\frac{a+b}{2}, G=\sqrt{a b}, H=\frac{2 a b}{a+b}$
i. A. $\mathrm{H}=\frac{\mathrm{a}+\mathrm{b}}{2} \cdot \frac{2 \mathrm{ab}}{\mathrm{a}+\mathrm{b}}=\mathrm{ab}=\mathrm{G}^{2}$
i.e. $\mathrm{G}^{2}=$ A.H
$\mathrm{G}$ is the geometric mean between $\mathrm{A} \& \mathrm{H}$.
Again $A-G=\frac{a+b}{2}-\sqrt{a b}=\frac{(\sqrt{a}-\sqrt{b})^{2}}{2}>0$
$\Rightarrow \mathrm{A}>\mathrm{G}$
Also $\mathrm{G}^{2}=\mathrm{A} . \mathrm{H}$
$ \begin{aligned} & \frac{\mathrm{G}}{\mathrm{H}}=\frac{\mathrm{A}}{\mathrm{G}}>1 \\ & \Rightarrow \frac{\mathrm{G}}{\mathrm{H}}>1 \text { or } \mathrm{G}>\mathrm{H} \end{aligned} $
Combining, $\mathrm{A}>\mathrm{G}>\mathrm{H}$
Note: If the numbers are equal, then $A=G=H$. Thus, $A \geq G \geq H$, equality holds when the numbers are equal.
ii. The equation with $a$ and $b$ as its roots is $\mathrm{x}^{2}-2 \mathrm{Ax}+\mathrm{G}^{2}=0$.
or if $A \& G$ be the A.M and G.M between two positive numbers $a \& b$ then $\frac{a}{b}=\frac{A+\sqrt{A^{2}-G^{2}}}{A-\sqrt{A^{2}-G^{2}}}$
iii. If A, G, H be the A.M, G.M, and H.M between three given numbers, a, b and c, then the equation having $a, b, c$ as its roots is
$\mathrm{x}^{3}-3 A \mathrm{x}^{2}+\frac{3 \mathrm{G}^{3}}{\mathrm{H}} \mathrm{x}-\mathrm{G}^{3}=0$
Proof: $A=\frac{a+b+c}{3} \Rightarrow a+b+c=3 A$
$ G=(a b c)^{1 / 3} \Rightarrow G^{3}=a b c $
$ \begin{aligned} & \frac{1}{\mathrm{H}}=\frac{\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}+\frac{1}{\mathrm{c}}}{3} \Rightarrow \frac{\mathrm{ab}+\mathrm{bc}+\mathrm{ca}}{3 \mathrm{abc}}=\frac{1}{\mathrm{H}} \\ & \text { or } \mathrm{ab}+\mathrm{bc}+\mathrm{ca}=\frac{3 \mathrm{abc}}{\mathrm{H}}=\frac{3 \mathrm{G}^{3}}{\mathrm{H}} \end{aligned} $
Equation having $\mathrm{a}, \mathrm{b}, \mathrm{c}$ as roots is
$ \begin{aligned} & x^{3}-(a+b+c) x+(a b+b c+c a) x^{2}-a b c=0 \\ & x^{3}-3 A x+\frac{3 G^{3}}{H} x-G^{3}=0 \end{aligned} $
Example : For distinct positive numbers $x, y, z$, prove that $(x+y)(y+z)(z+x)>8 x y z$
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Solution: We have $\mathrm{A}>\mathrm{G}$
For positive numbers $\mathrm{x}$ and $\mathrm{y}, \mathrm{x}+\mathrm{y}>2 \sqrt{\mathrm{xy}}$
For positive numbers y and $\mathrm{z}, \mathrm{y}+\mathrm{z}>2 \sqrt{\mathrm{yz}}$
For positive numbers $\mathrm{z}$ and $\mathrm{x}, \mathrm{z}+\mathrm{x}>2 \sqrt{\mathrm{zx}}$
Multiplying,
$ (x+y)(y+z)(z+x)>8 \sqrt{x^{2} y^{2} z^{2}} $
i.e.,
$ (x+y)(y+z)(z+x)>8 x y z $
Solved examples
1. If $a, b, c$ are positive then prove that $((1+a)(1+b)(1+c))^{7}>7^{7} a^{4} b^{4} c^{4}$.
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Solution: $(1+a)(1+b)(1+c)=1+a+b+c+a b+b c+c a+a b c$
$ \begin{aligned} & >a+b+c+a b+b c+c a+a b c \\ & \geq 7(a \cdot b \cdot c \cdot a b \cdot b c \cdot c a \cdot a b c)^{1 / 7} \quad(\because A \geq G) \end{aligned} $
i.e. $\quad(1+a)(1+b)(1+c) \geq 1+7\left(a^{4} b^{4} c^{4}\right)^{1 / 7}>7\left(a^{4} b^{4} c^{4}\right)^{1 / 7}$
$ ((1+a)(1+b)(1+c))^{7}>7^{7} a^{4} b^{4} c^{4} $
2. Maximum value of $x y z$ for positive values of $x, y, z$ if $y z+z x+x y=12$ is
(a). $64$
(b). $4^{3 / 2}$
(c). $8$
(d). none of these
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Solution: Apply $ A \geq G$ for $y z, z x \& x y$
$ \frac{y z+z x+x y}{3}>\left(x^{2} y^{2} z^{2}\right)^{1 / 3} $
$ \begin{aligned} & \frac{12}{3} \geq(\mathrm{xyz})^{2 / 3} \\ & (\mathrm{xyz}) \leq 4^{3 / 2} \\ & \mathrm{xyz} \leq 8 \end{aligned} $
Answer: c
3. Maximum value of $x^{2} y^{3}$ where $x$ & $y$ lie in $1^{\text {st }}$ quadrant in the line $3 x+4 y=5$.
(a). $\frac{5}{16}$
(b). $\frac{3}{8}$
(c). $\frac{5}{8}$
(d). $\frac{3}{16}$
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Solution: $x^{2} y^{3}=x \cdot x \cdot y \cdot y \cdot y$
$3x+4 y=\frac{3 x}{2}+\frac{3 x}{2}+\frac{4 y}{3}+\frac{4 y}{3}+\frac{4 y}{3}$
$\mathrm{A} \geq \mathrm{G}$
$ \frac{\frac{3 x}{2}+\frac{3 x}{2}+\frac{4 y}{3}+\frac{4 y}{3}+\frac{4 y}{3}}{5} \geq\left(\frac{3 x}{2} \cdot \frac{3 x}{2} \cdot \frac{4 y}{3} \cdot \frac{4 y}{3} \cdot \frac{4 y}{3}\right)^{1 / 5} $
$ \frac{5}{5} \geq\left(\frac{16 x^{2} y^{3}}{3}\right)^{\frac{1}{5}} $
$\Rightarrow \quad x^{2} y^{3} \leq \frac{3}{16}$
Answer: d
4. If $\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}=1=\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}$, then maximum value of $\mathrm{ax}+\mathrm{by}+\mathrm{cz}$ is $(\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{x}, \mathrm{y}, \mathrm{z}$ are positive real numbers)
(a). 4
(b). 3
(c). 2
(d). 1
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Solution: $A>G$
$ \frac{\mathrm{a}^{2}+\mathrm{x}^{2}}{2}>\sqrt{\mathrm{a}^{2} \mathrm{x}^{2}} \Rightarrow \quad \mathrm{a}^{2}+\mathrm{x}^{2}>2 \mathrm{ax} $
Similarly
$ \begin{aligned} & \mathrm{b}^{2}+\mathrm{y}^{2}>2 \mathrm{by} \\ & \mathrm{c}^{2}+\mathrm{z}^{2}>2 \mathrm{cz} \end{aligned} $
adding, $\left(a^{2}+b^{2}+c^{2}\right)+\left(x^{2}+y^{2}+z^{2}\right)>2(a x+b y+c z)$
$\Rightarrow \quad \mathrm{ax}+\mathrm{by}+\mathrm{cz}<\frac{1+1}{2}$
Answer: d
5. If $a, b, c$ are positive then the minimum value of $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$ is
(a). $\frac{2}{3}$
(b). $\frac{3}{2}$
(c). 1
(d). none of these
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Solution: Apply $A \geq H$ for $\frac{1}{\mathrm{~b}+\mathrm{c}}, \frac{1}{\mathrm{c}+\mathrm{a}}, \frac{1}{\mathrm{a}+\mathrm{b}}$
$ \begin{aligned} & \frac{\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}}{3} \geq \frac{3}{a+b+b+c+c+a} \\ \Rightarrow \quad & \frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b} \geq \frac{9}{2(a+b+c)} \\ \Rightarrow \quad & \frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b} \geq \frac{9}{2} \\ \Rightarrow \quad & \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+3 \geq \frac{9}{2} \\ \Rightarrow \quad & \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geq \frac{3}{2} \end{aligned} $
Answer: b
6. If $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are three positive numbers, then $(\mathrm{x}+\mathrm{y}+\mathrm{z})\left(\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}\right)>………….$
(a). $3$
(b). $9$
(c). $\frac{1}{3}$
(d). none of these
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Solution: $\quad \frac{x+y+z}{3}>(x y z)^{1 / 3}$
Also $\frac{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}{3}>\left(\frac{1}{x} \cdot \frac{1}{y} \cdot \frac{1}{z}\right)^{1 / 3}$ $(\because A>G)$
Multiplying, $\quad \frac{(\mathrm{x}+\mathrm{y}+\mathrm{z})\left(\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}\right)}{9}>1$
Answer: b (or apply $\mathrm{A} \geq \mathrm{H}$ for $\mathrm{x}, \mathrm{y}, \mathrm{z}$ to get the result)
7. Prove that: ${ }^{n} C _{1} \cdot\left({ }^{n} C _{2}\right)^{2}\left({ }^{n} C _{3}\right)^{3} \ldots \ldots . .\left({ }^{n} C _{n}\right)^{n} \leq\left(\frac{2^{n}}{n+1}\right)$
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Solution: Let $\mathrm{S}={ }^{n} \mathrm{C} _{1}+2^{\mathrm{n}} \mathrm{C} _{2}+3^{\mathrm{n}} \mathrm{C} _{3}+$ n ${ }^{n} C _{n}$
$ =\sum _{r=1}^{n} \quad r \cdot{ }^{n} C _{r}=\sum _{r=1}^{n} n \cdot{ }^{n-1} C _{r-1}=n 2^{n-1} $
Now $A \geq G$
$\frac{{ }^{n} C _{1}+\left({ }^{n} C _{2}+{ }^{n} C _{2}\right)+\left({ }^{n} C _{3}+{ }^{n} C _{3}+{ }^{n} C _{3}\right)+\ldots . .\left({ }^{n} C _{n}+{ }^{n} C _{n}+\ldots \ldots+{ }^{n} C _{n}\right)}{1+2+3+\ldots \ldots \ldots .+n}$
$\geq\left({ }^{n} C _{1}\left({ }^{n} C _{2}\right)^{2} \ldots \ldots . .\left({ }^{n} C _{n}\right)^{n}\right)^{\frac{1}{1+2+\ldots .+n}}$
$\Rightarrow \quad \frac{\mathrm{n} \cdot 2^{\mathrm{n}-1}}{\frac{\mathrm{n}(\mathrm{n}+1)}{2}} \geq\left({ }^{\mathrm{n}} \mathrm{C} _{1}\left({ }^{\mathrm{n}} \mathrm{C} _{2}\right)^{2} \ldots \ldots . .\left({ }^{\mathrm{n}} \mathrm{C} _{\mathrm{n}}\right)^{\mathrm{n}}\right)^{\frac{2}{(\mathrm{n}+1)}}$
$\Rightarrow \quad\left({ }^{n} C _{1} \cdot\left({ }^{n} C _{2}\right)^{2}\left({ }^{n} C _{3}\right)^{3} \ldots \ldots .\left({ }^{n} C _{n}\right)^{n}\right) \leq\left(\frac{2^{n}}{n+1}\right)^{\frac{n(n+1)}{2}}=\left(\frac{2^{n}}{n+1}\right)^{n+1} C _{2}$
8. If $a, b, c, d$ are in H.P. then
(a). $ \mathrm{a}+\mathrm{d}>\mathrm{b}+\mathrm{c}$
(b). $a d>b c$
(c). $a d=b c$
(d). none of these
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Solution: $ \mathrm{a}, \mathrm{b}, \mathrm{c}$ are in H.P $\Rightarrow \mathrm{b}$ is the H.M of $\mathrm{a} \& \mathrm{c}$.
A.M of $\mathrm{a} \& \mathrm{c}=\frac{\mathrm{a}+\mathrm{c}}{2}$
we have $\mathrm{A}>\mathrm{H} \Rightarrow \frac{\mathrm{a}+\mathrm{c}}{2}>\mathrm{b} \Rightarrow \mathrm{a}+\mathrm{c}>2 \mathrm{~b}$
Similarly $ b+d>2 c$
Adding, $ \mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}>2 \mathrm{~b}+2 \mathrm{c} \quad \Rightarrow \quad \mathrm{a}+\mathrm{d}>\mathrm{b}+\mathrm{c}$
Also $a, b, c$ are in H.P. $\Rightarrow b$ is the H.M of $a \& c$
G.M of $\mathrm{a} \& \mathrm{c}=\sqrt{\mathrm{ac}}$
$\mathrm{G}>\mathrm{H} \Rightarrow \sqrt{\mathrm{ac}}>\mathrm{b}$
Similarly $\sqrt{\mathrm{b}} \mathrm{d}>\mathrm{c}$
Multiplying, $\sqrt{\mathrm{abcd}}>\mathrm{bc} \quad \Rightarrow \quad \mathrm{ad}>\mathrm{bc}$
Answer: a, b
Practice questions
1. If $\mathrm{a}+\mathrm{b}+\mathrm{c}=1$, then find $k$ such that
$\frac{\mathrm{k}}{27 \mathrm{abc}}>\left(\frac{1}{\mathrm{a}}-1\right)\left(\frac{1}{\mathrm{~b}}-1\right)\left(\frac{1}{\mathrm{c}}-1\right)>\mathrm{k}$
(a). 8
(b). 7
(c). 3
(d). none of these
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Answer: (a)2. A rod of fixed length $k$ slides along the coordinate axes. If it meets the axes at $A(a, 0)$ and $B(0, b)$, then the minimum value of $\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}$ is
(a). $0$
(b). $8$
(c). $\mathrm{k}^{2}-4+\frac{4}{\mathrm{k}^{2}}$
(d). $\mathrm{k}^{2}+4+\frac{4}{\mathrm{k}^{2}}$
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Answer: (d)3. If positive numbers $a, b, c$ be in H.P, the equation $x^{2}-k x+2 b^{101}-a^{101}-c^{101}=0(k \varepsilon R)$ has
(a). both roots imaginary
(b). one root is positive and other is negative
(c). both roots positive
(d). both roots negative
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Answer: (b)4. If $\mathrm{n} \varepsilon \mathrm{N}, \mathrm{n}^{\mathrm{n}}\left(\frac{\mathrm{n}+1}{2}\right)^{2 \mathrm{n}}>\mathrm{k}$ where $\mathrm{k}$ is
(a). $(2 n !)^{3}$
(b). $2(n !)^{3}$
(c). $(n !)^{3}$
(d). none of these
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Answer: (c)5. If $x, y, z \in R^{+}$, then is $\frac{y z}{y+z}+\frac{x z}{x+z}+\frac{x y}{x+y}$ is always
(a). $\leq \frac{1}{2}(x+y+z)$
(b). $\geq \frac{1}{3} \sqrt{\mathrm{xyz}}$
(c). $\leq \frac{1}{3}(\mathrm{x}+\mathrm{y}+\mathrm{z})$
(d). $\geq \frac{1}{2} \sqrt{\mathrm{xyz}}$
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Answer: (a)6. $\sum _{\mathrm{i}=0}^{\infty} \sum _{\mathrm{j}=0}^{\infty} \sum _{\mathrm{k}=0}^{\infty} \frac{1}{3^{\mathrm{i}} 3^{\mathrm{j}} 3^{\mathrm{k}}}$ is
$(\mathrm{i} \neq \mathrm{j} \neq \mathrm{k})$
(a). $\frac{1}{27}$
(b). $\frac{81}{208}$
(c). $1$
(d). none of these
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Answer: (b)7. Minimum value of $\frac{(\mathrm{x}-1)(\mathrm{x}-2)}{\mathrm{x}-3}: \forall \mathrm{x}>3$ is
(a). $3 \sqrt{2}-2$
(b). $3+2 \sqrt{2}$
(c). $3+2 \sqrt{3}$
(d). $3 \sqrt{2}+2$
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Answer: (b)8. The least value of $6 \tan ^{2} \theta+54 \cot ^{2} \theta+18$ is
i. 54 when A.M. $\geq$ G.M is applied for $6 \tan ^{2} \theta, 54 \cot ^{2} \theta, 18$
ii. 54 when $\mathrm{A} . \mathrm{M} \geq \mathrm{G} . \mathrm{M}$ is applied for $6 \tan ^{2} \theta, 54 \cot ^{2} \theta$ and 18 is added further.
iii. 78 when $\tan ^{2} \theta=\cot ^{2} \theta$.
(a). (iii) is correct
(b). (i) is correct (ii) is flase
(c). (i) and (ii) are correct
(d). none of these
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Answer: (c)9. If A, G, H are A.M, G.M, H.M between the same two numbers, such that $\mathrm{A}-\mathrm{G}=15$ and $\mathrm{A}-\mathrm{H}=27$, then the numbers are
(a). 100, 50
(b). 120, 30
(c). 90, 60
(d). none of these
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Answer: (b)10. If $a, b, c \varepsilon R$, the square root of $a^{2}+b^{2}+c^{2}-a b-b c-a c$ is greater than or equal to
(a). $\frac{\sqrt{3}}{2} \max \{|\mathbf{b}-\mathrm{c}|,|\mathrm{c}-\mathrm{a}|,|\mathrm{a}-\mathrm{b}|\}$
(b). $\frac{3}{2} \max \{|\mathrm{b}-\mathrm{c}|,|\mathrm{c}-\mathrm{a}|,|\mathrm{a}-\mathrm{b}|\}$
(c). $\max \{|\mathrm{b}-\mathrm{c}|,|\mathrm{c}-\mathrm{a}|,|\mathrm{a}-\mathrm{b}|\}$
(d). $\frac{\sqrt{3}}{4} \max \{|\mathbf{b}-\mathrm{c}|,|\mathrm{c}-\mathrm{a}|,|\mathrm{a}-\mathrm{b}|\}$
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Answer: (a)11. If $\mathrm{x} _{1}, \mathrm{x} _{2}, \mathrm{x} _{3}, \mathrm{x} _{4}$ are four positive real numbers such that $\mathrm{x} _{1}+\frac{1}{\mathrm{x} _{2}}=4, \mathrm{x} _{2}+\frac{1}{\mathrm{x} _{3}}=1, \mathrm{x} _{3}+\frac{1}{\mathrm{x} _{4}}=4, \mathrm{x} _{4}+\frac{1}{\mathrm{x} _{1}}=1$ then
(a). $\mathrm{x} _{1}=\mathrm{x} _{3}$ and $\mathrm{x} _{2}=\mathrm{x} _{4}$
(b). $\mathrm{x} _{2}=\mathrm{x} _{4}$ but $\mathrm{x} _{1} \neq \mathrm{x} _{3}$
(c). $\mathrm{x} _{1} \mathrm{x} _{2}=1, \mathrm{x} _{3} \mathrm{x} _{4}=-1$
(d). $\mathrm{x} _{3} \mathrm{x} _{4}=1, \mathrm{x}, \mathrm{x} _{2} \neq 1$
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Answer: (a)12. If $\mathrm{a}, \mathrm{b}, \mathrm{c}>0$ and $\mathrm{a}(1-\mathrm{b})>\frac{1}{4}, \mathrm{~b}(1-\mathrm{c})>\frac{1}{4}, \mathrm{c}(1-\mathrm{a})>\frac{1}{4}$, then
(a). never possible
(b). always true
(c). cannot be discussed
(d). none of these
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Answer: (a)13. If $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are the sides of a triangle, then $\frac{1}{\mathrm{~b}+\mathrm{c}}, \frac{1}{\mathrm{c}+\mathrm{a}}, \frac{1}{\mathrm{a}+\mathrm{b}}$ are also the sides of the triangle is
(a). sometimes true
(b). always true
(c). cannot be discussed
(d). never true
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Answer: (b)14. Given $\mathrm{n}^{4}<10^{\mathrm{n}}$ for a fixed positive integer $\mathrm{n} \geq 2$, then
(a). $(\mathrm{n}+1)^{4}<10^{\mathrm{n}+1}$
(b). $(\mathrm{n}+1)^{4}>10^{\mathrm{n}+1}$
(c). nothing can be said
(d). none of these
