Question 54
If $$a^{2} + b^{2} + \dfrac{1}{a^{2}} + \dfrac{1}{b^{2}} = 4$$ then the value of $$a^{2} + b^{2}$$
Solution
Let's rewrite the equation:
$$a^2+\dfrac{1}{a^2}+b^2+\dfrac{1}{b^2}=4$$
We know that $$x^2+\dfrac{1}{x^2}\ge\ 2$$
If we compare, we get,
$$a^2+\dfrac{1}{a^2}=\ 2$$ and $$b^2+\dfrac{1}{b^2}=\ 2$$
This is possible when $$a^2=1\ \&\ b^2=1$$
Hence, $$a^2+b^2\ =\ 1\ +\ 1\ =\ 2$$
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