Hello Ernesto,
If $b+c,\,a+c,\,a+b$ are in harmonic progression, then their reciprocals are in arithmetic progression, and we may state:
(i) $$\frac{1}{a+c}=\frac{1}{b+c}+d$$
Multiply through by $$(a+c)(b+c)$$ to obtain:
$$b+c=a+c+d(a+c)(b+c)$$
Subtract $c$ from both sides and expand the product on the far right:
$$b=a+d(ab+ac+bc+c^2)$$
Distribute on the right:
$$b=a+abd+acd+bcd+c^2d$$
Subtract through by $$a+abd+acd+bcd$$:
$$b-a-abd-acd-bcd=c^2d$$
Divide through by $d$ and arrange as:
$$c^2=\frac{b-a-abd-acd-bcd}{d}$$
(ii) $$\frac{1}{a+b}=\frac{1}{b+c}+2d$$
Multiply through by $$(a+b)(b+c)$$ to obtain:
$$b+c=a+b+2d(a+b)(b+c)$$
Subtract $b$ from both sides and expand the product on the far right:
$$c=a+2d(ab+ac+b^2+bc)$$
Distribute on the right:
$$c=a+2abd+2acd+2b^2d+2bcd$$
Subtract through by $$a+2abd+2acd+2bcd$$:
$$c-a-2abd-2acd-2bcd=2b^2d$$
Divide through by $2d$ and arrange as:
$$b^2=\frac{c-a-2abd-2acd-2bcd}{2d}$$
(iii) $$\frac{1}{a+b}=\frac{1}{a+c}+d$$
Multiply through by $$(a+b)(a+c)$$ to obtain:
$$a+c=a+b+d(a+b)(a+c)$$
Subtract $a$ from both sides and expand the product on the far right:
$$c=b+d(a^2+ac+ab+bc)$$
Distribute on the right:
$$c=b+a^2d+acd+abd+bcd$$
Subtract through by $$b+acd+abd+bcd$$:
$$c-b-acd-abd-bcd=a^2d$$
Divide through by $d$ and arrange as:
$$a^2=\frac{c-b-acd-abd-bcd}{d}$$
Now in order for $a^2,\,b^2,\,c^2$ to be in arithmetic progression, we require:
$$b^2-a^2=c^2-b^2$$
Add through by $$a^2+b^2$$:
$$2b^2=a^2+c^2$$
Substitute for $a^2,\,b^2,\,c^2$ the expressions we found above:
$$2\left(\frac{c-a-2abd-2acd-2bcd}{2d} \right)=\frac{c-b-acd-abd-bcd}{d}+\frac{b-a-abd-acd-bcd}{d}$$
Distributing the 2 on the left, and the multiplying through by $d$, we find:
$$c-a-2abd-2acd-2bcd=c-b-acd-abd-bcd+b-a-abd-acd-bcd$$
Add through by $$2abd+2acd+2bcd$$:
$$c-a=c-b+b-a$$
Collect like terms:
$$c-a=c-a$$
Subtract through by $$c-a$$:
$$0=0$$
This is an identity, which proves that given $b+c,\,a+c,\,a+b$ are in harmonic progression, then $a^2,\,b^2,\,c^2$ must be in arithmetic progression.
