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Hint: First, We should try to convert given terms ${a^2}$, ${b^2}$, ${c^2}$ to the terms somewhat similar to terms what is to be shown i.e. $b + c$,$c + a$,$a + b$ by adding or subtracting some additional terms, see if some factors are made or not.

Given ${a^2},{b^2},{c^2}$ are in A.P.

By adding $(ab + ac + bc)$ to each term of given A.P.

We see that ${a^2} + ab + ac + bc$,${b^2} + ab + ac + bc$,\[{c^2} + ab + ac + bc\] are also in A.P.

Convert each term in factor form

${a^2} + ab + ac + bc = a\left( {a + b} \right) + c\left( {a + b} \right) = \left( {a + c} \right)\left( {a + b} \right)$

${b^2} + ab + ac + bc = b\left( {b + a} \right) + c\left( {a + b} \right) = \left( {a + b} \right)\left( {b + c} \right)$

${c^2} + ac + ab + bc = c\left( {c + a} \right) + b\left( {a + c} \right) = \left( {c + b} \right)\left( {a + c} \right)$

Also we can write terms of above A.P. in another way like

$\left( {a + b} \right)\left( {a + c} \right)$,$\left( {a + b} \right)\left( {b + c} \right)$,$\left( {c + a} \right)\left( {c + b} \right)$ are in A.P.

Now we divide each term by $\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)$

We get$\dfrac{1}{{b + c}}$,$\dfrac{1}{{c + a}}$,$\dfrac{1}{{a + b}}$ are in A.P.

So we can say that$b + c$,$c + a$,$a + b$ are in H.P.

Hence proved

Note: Arithmetic progression (A.P) means a sequence in which each differs from the preceding one by a constant quantity. Harmonic progression means a series when their reciprocal is in arithmetic progression.

Given ${a^2},{b^2},{c^2}$ are in A.P.

By adding $(ab + ac + bc)$ to each term of given A.P.

We see that ${a^2} + ab + ac + bc$,${b^2} + ab + ac + bc$,\[{c^2} + ab + ac + bc\] are also in A.P.

Convert each term in factor form

${a^2} + ab + ac + bc = a\left( {a + b} \right) + c\left( {a + b} \right) = \left( {a + c} \right)\left( {a + b} \right)$

${b^2} + ab + ac + bc = b\left( {b + a} \right) + c\left( {a + b} \right) = \left( {a + b} \right)\left( {b + c} \right)$

${c^2} + ac + ab + bc = c\left( {c + a} \right) + b\left( {a + c} \right) = \left( {c + b} \right)\left( {a + c} \right)$

Also we can write terms of above A.P. in another way like

$\left( {a + b} \right)\left( {a + c} \right)$,$\left( {a + b} \right)\left( {b + c} \right)$,$\left( {c + a} \right)\left( {c + b} \right)$ are in A.P.

Now we divide each term by $\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)$

We get$\dfrac{1}{{b + c}}$,$\dfrac{1}{{c + a}}$,$\dfrac{1}{{a + b}}$ are in A.P.

So we can say that$b + c$,$c + a$,$a + b$ are in H.P.

Hence proved

Note: Arithmetic progression (A.P) means a sequence in which each differs from the preceding one by a constant quantity. Harmonic progression means a series when their reciprocal is in arithmetic progression.

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