Question
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,c are in H.P. and they are distinct and positive, then prove that an+cn>2bn
Solution
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 a1+c1=b2⇒b=a+c2ac
Consider geometric mean of a and c = ac
Harmonic mean of a and c = a+c2ac
We know that G.M > H.P
So, we have
Raising its power to n on both sides, we get
⇒(ac)n>bn ⇒(ac)2n>bn....................................(1)Let an and cn be two numbers. Since a,c are positive and distinct, an and cn are also positive.
We know that for two numbers A.M > G.M > H.M
Now, Arithmetic mean (A.M) of an and cn = 2an+cn
Geometric mean (G.M) of an and cn = ancn=(ac)2n
Harmonic mean of an and cn = an+cn2ancn=an+cn2(ac)2n
Since A.M > G.M
Hence proved that an+cn>2bn
Note : Three numbers x,y,z are said to be in Harmonic progression if it satisfies the condition x1+z1=y2. Arithmetic mean, geometric mean and harmonic mean of two numbers x and y is given by 2x+y,xy,x+y2xy respectively.