# 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.

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.