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Question: If \[a,b,c\] are in H.P. and they are distinct and positive, then prove that \[{a^n} + {c^n} > 2{b^n...

If a,b,ca,b,c are in H.P. and they are distinct and positive, then prove that an+cn>2bn{a^n} + {c^n} > 2{b^n}

Explanation

Solution

First of all, find an equation between the numbers a,b,ca,b,c which is in H.P. Then find the geometric mean and harmonic mean of aa and cc. 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,ca,b,c are in H.P. and they are distinct and positive.
So, we have 1a+1c=2bb=2aca+c\dfrac{1}{a} + \dfrac{1}{c} = \dfrac{2}{b} \Rightarrow b = \dfrac{{2ac}}{{a + c}}
Consider geometric mean of aa and cc = ac\sqrt {ac}
Harmonic mean of aa and cc = 2aca+c\dfrac{{2ac}}{{a + c}}
We know that G.M > H.P
So, we have

ac>2aca+c ac>b [2aca+c=b]  \Rightarrow \sqrt {ac} > \dfrac{{2ac}}{{a + c}} \\\ \Rightarrow \sqrt {ac} > b\,{\text{ }}\left[ {\because \dfrac{{2ac}}{{a + c}} = b} \right] \\\

Raising its power to nn on both sides, we get

(ac)n>bn (ac)n2>bn....................................(1)  \Rightarrow {\left( {\sqrt {ac} } \right)^n} > {b^n} \\\ \Rightarrow {\left( {ac} \right)^{\dfrac{n}{2}}} > {b^n}....................................\left( 1 \right) \\\

Let an{a^n} and cn{c^n} be two numbers. Since a,ca,c are positive and distinct, an{a^n} and cn{c^n} are also positive.
We know that for two numbers A.M > G.M > H.M
Now, Arithmetic mean (A.M) of an{a^n} and cn{c^n} = an+cn2\dfrac{{{a^n} + {c^n}}}{2}
Geometric mean (G.M) of an{a^n} and cn{c^n} = ancn=(ac)n2\sqrt {{a^n}{c^n}} = {\left( {ac} \right)^{\dfrac{n}{2}}}
Harmonic mean of an{a^n} and cn{c^n} = 2ancnan+cn=2(ac)n2an+cn\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

an+cn2>(ac)n2 an+cn>2(ac)n2 an+cn>2bn [equation (1)]  \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 an+cn>2bn{a^n} + {c^n} > 2{b^n}

Note : Three numbers x,y,zx,y,z are said to be in Harmonic progression if it satisfies the condition 1x+1z=2y\dfrac{1}{x} + \dfrac{1}{z} = \dfrac{2}{y}. Arithmetic mean, geometric mean and harmonic mean of two numbers xx and yy is given by x+y2,xy,2xyx+y\dfrac{{x + y}}{2},\sqrt {xy} ,\dfrac{{2xy}}{{x + y}} respectively.