If \[a,b,c\] are in H.P. and they are distinct and positive, then prove that \[{a^n} + {c^n} > 2{b^n}\]
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Hint: First of all, find an equation between the numbers \[a,b,c\] which is in H.P. Then find the geometric mean and harmonic mean of \[a\] and \[c\]. Use A.M > G.M > H.P to prove the given statement. So, use this concept to reach the solution of the given problem.
Complete step-by-step answer:
Given that \[a,b,c\] are in H.P. and they are distinct and positive.
So, we have \[\dfrac{1}{a} + \dfrac{1}{c} = \dfrac{2}{b} \Rightarrow b = \dfrac{{2ac}}{{a + c}}\]
Consider geometric mean of \[a\] and \[c\] = \[\sqrt {ac} \]
Harmonic mean of \[a\] and \[c\] = \[\dfrac{{2ac}}{{a + c}}\]
We know that G.M > H.P
So, we have
\[
\Rightarrow \sqrt {ac} > \dfrac{{2ac}}{{a + c}} \\
\Rightarrow \sqrt {ac} > b\,{\text{ }}\left[ {\because \dfrac{{2ac}}{{a + c}} = b} \right] \\
\]
Raising its power to \[n\] on both sides, we get
\[
\Rightarrow {\left( {\sqrt {ac} } \right)^n} > {b^n} \\
\Rightarrow {\left( {ac} \right)^{\dfrac{n}{2}}} > {b^n}....................................\left( 1 \right) \\
\]
Let \[{a^n}\] and \[{c^n}\] be two numbers. Since \[a,c\] are positive and distinct, \[{a^n}\] and \[{c^n}\] are also positive.
We know that for two numbers A.M > G.M > H.M
Now, Arithmetic mean (A.M) of \[{a^n}\] and \[{c^n}\] = \[\dfrac{{{a^n} + {c^n}}}{2}\]
Geometric mean (G.M) of \[{a^n}\] and \[{c^n}\] = \[\sqrt {{a^n}{c^n}} = {\left( {ac} \right)^{\dfrac{n}{2}}}\]
Harmonic mean of \[{a^n}\] and \[{c^n}\] = \[\dfrac{{2{a^n}{c^n}}}{{{a^n} + {c^n}}} = \dfrac{{2{{\left( {ac} \right)}^{\dfrac{n}{2}}}}}{{{a^n} + {c^n}}}\]
Since A.M > G.M
\[
\Rightarrow \dfrac{{{a^n} + {c^n}}}{2} > {\left( {ac} \right)^{\dfrac{n}{2}}} \\
\Rightarrow {a^n} + {c^n} > 2{\left( {ac} \right)^{\dfrac{n}{2}}} \\
\therefore {a^n} + {c^n} > 2{b^n}{\text{ }}\left[ {\because {\text{equation }}\left( 1 \right)} \right] \\
\]
Hence proved that \[{a^n} + {c^n} > 2{b^n}\]
Note: Three numbers \[x,y,z\] are said to be in Harmonic progression if it satisfies the condition \[\dfrac{1}{x} + \dfrac{1}{z} = \dfrac{2}{y}\]. Arithmetic mean, geometric mean and harmonic mean of two numbers \[x\] and \[y\] is given by \[\dfrac{{x + y}}{2},\sqrt {xy} ,\dfrac{{2xy}}{{x + y}}\] respectively.
Complete step-by-step answer:
Given that \[a,b,c\] are in H.P. and they are distinct and positive.
So, we have \[\dfrac{1}{a} + \dfrac{1}{c} = \dfrac{2}{b} \Rightarrow b = \dfrac{{2ac}}{{a + c}}\]
Consider geometric mean of \[a\] and \[c\] = \[\sqrt {ac} \]
Harmonic mean of \[a\] and \[c\] = \[\dfrac{{2ac}}{{a + c}}\]
We know that G.M > H.P
So, we have
\[
\Rightarrow \sqrt {ac} > \dfrac{{2ac}}{{a + c}} \\
\Rightarrow \sqrt {ac} > b\,{\text{ }}\left[ {\because \dfrac{{2ac}}{{a + c}} = b} \right] \\
\]
Raising its power to \[n\] on both sides, we get
\[
\Rightarrow {\left( {\sqrt {ac} } \right)^n} > {b^n} \\
\Rightarrow {\left( {ac} \right)^{\dfrac{n}{2}}} > {b^n}....................................\left( 1 \right) \\
\]
Let \[{a^n}\] and \[{c^n}\] be two numbers. Since \[a,c\] are positive and distinct, \[{a^n}\] and \[{c^n}\] are also positive.
We know that for two numbers A.M > G.M > H.M
Now, Arithmetic mean (A.M) of \[{a^n}\] and \[{c^n}\] = \[\dfrac{{{a^n} + {c^n}}}{2}\]
Geometric mean (G.M) of \[{a^n}\] and \[{c^n}\] = \[\sqrt {{a^n}{c^n}} = {\left( {ac} \right)^{\dfrac{n}{2}}}\]
Harmonic mean of \[{a^n}\] and \[{c^n}\] = \[\dfrac{{2{a^n}{c^n}}}{{{a^n} + {c^n}}} = \dfrac{{2{{\left( {ac} \right)}^{\dfrac{n}{2}}}}}{{{a^n} + {c^n}}}\]
Since A.M > G.M
\[
\Rightarrow \dfrac{{{a^n} + {c^n}}}{2} > {\left( {ac} \right)^{\dfrac{n}{2}}} \\
\Rightarrow {a^n} + {c^n} > 2{\left( {ac} \right)^{\dfrac{n}{2}}} \\
\therefore {a^n} + {c^n} > 2{b^n}{\text{ }}\left[ {\because {\text{equation }}\left( 1 \right)} \right] \\
\]
Hence proved that \[{a^n} + {c^n} > 2{b^n}\]
Note: Three numbers \[x,y,z\] are said to be in Harmonic progression if it satisfies the condition \[\dfrac{1}{x} + \dfrac{1}{z} = \dfrac{2}{y}\]. Arithmetic mean, geometric mean and harmonic mean of two numbers \[x\] and \[y\] is given by \[\dfrac{{x + y}}{2},\sqrt {xy} ,\dfrac{{2xy}}{{x + y}}\] respectively.