Question

If a, b, c, d are non zero real numbers such that
$$\left( a^{2}+b^{2}+c^{2}\right) \left( b^{2}+c^{2}+d^{2}\right) \leq \left( ab+bc+cd\right)^{2} $$, then a, b, c, d are in
A) AP
B) GP
C) HP
D) none of these

Answer 1

Hint: In this question it is given that if a, b, c, d are non zero real numbers such that
$$\left( a^{2}+b^{2}+c^{2}\right) \left( b^{2}+c^{2}+d^{2}\right) \leq \left( ab+bc+cd\right)^{2} $$, then we have to find the relation of a, b, c and d. So to find the solution we have to expand and rearrange the given inequation by using proper identity.
So the identity that we will be using are,
$$\left( x+y+z\right)^{2} =x^{2}+y^{2}+z^{2}+2xy+2yz+2zx$$.......(1)
$$\left( x-y\right)^{2} =x^{2}-2xy+y^{2}$$..........(2)
Complete step-by-step solution:
Given,
$$\left( a^{2}+b^{2}+c^{2}\right) \left( b^{2}+c^{2}+d^{2}\right) \leq \left( ab+bc+cd\right)^{2} $$
$$\Rightarrow a^{2}b^{2}+a^{2}c^{2}+a^{2}d^{2}+b^{4}+b^{2}c^{2}+b^{2}d^{2}+c^{2}b^{2}+c^{4}+c^{2}d^{2}\leq \left( ab\right)^{2} +\left( bc\right)^{2} +\left( cd\right)^{2} +2\left( ab\right) \left( bc\right) +2\left( bc\right) \left( cd\right) +2\left( cd\right) \left( ab\right) $$ [ using formula (1) in RHS]
$$\Rightarrow a^{2}b^{2}+a^{2}c^{2}+a^{2}d^{2}+b^{4}+b^{2}c^{2}+b^{2}d^{2}+c^{2}b^{2}+c^{4}+c^{2}d^{2}\leq a^{2}b^{2}+b^{2}c^{2}+c^{2}d^{2}+2acb^{2}+2bdc^{2}+2adbc$$
Now, cancelling $a^{2}b^{2},\ b^{2}c^{2},\ c^{2}d^{2}$ in the both side of the above equation, we get
$$a^{2}c^{2}+a^{2}d^{2}+b^{4}+b^{2}d^{2}+c^{2}b^{2}+c^{4}\leq 2acb^{2}+2bdc^{2}+2adbc$$
$$\Rightarrow a^{2}c^{2}-2acb^{2}+b^{4}+b^{2}d^{2}-2bdc^{2}+c^{4}+a^{2}d^{2}-2adbc+b^{2}c^{2}\leq 0$$ [ by changing the side]
$$\Rightarrow \{ \left( ac\right)^{2} -2\left( ac\right) b^{2}+\left( b^{2}\right)^{2} \} +\{ \left( bd\right)^{2} -2\left( bd\right) c^{2}+\left( c^{2}\right)^{2} \} +\{ \left( ad\right)^{2} -2\left( ad\right) \left( bc\right) +\left( bc\right)^{2} \} \leq 0$$
$$\Rightarrow \left( ac-b^{2}\right)^{2} +\left( bd-c^{2}\right)^{2} +\left( ad-bc\right)^{2} \leq 0$$ [ using formula (2)]
If the summation of the squires of more than 1 term is zero then it implies each term is equal to zero.
i.e, $$\left( ac-b^{2}\right) =0,\ \left( bd-c^{2}\right) =0\text{ and } \left( ad-bc\right) =0$$
Therefore, $b^{2}=ac$; $c^{2}=bd$; $ad=bc$
Hence, a, b, c, d are in G.P.
Therefore, the correct option is option B.
Note: So you might be thinking how we concluded a, b, c, d are in GP just by these obtained expressions $b^{2}=ac$; $c^{2}=bd$; $ad=bc$, because as we know that if a, b, c, d are in Geometric Progression(GP) then ‘b’ and ‘c’ will be the geometric mean of a, c and b, d respectively, i.e, $b^{2}=ac$; $c^{2}=bd$, and they also have common ratio, i.e, $$\dfrac{b}{a} =\dfrac{c}{b} =\dfrac{d}{c}$$, so from here we can write $ad=bc$.