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  • <img alt="\\I\! \! f\; a_{1}, a_{2}, a_{3}, a_{4}>0, then \;the \;minimum\; value\; of\\\\ \left(a_{1}+a_{2}+a_{3}+a_{4}\right)\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4

\\I\! \! f\; a_{1}, a_{2}, a_{3}, a_{4}>0, then \;the \;minimum\; value\; of\\\\ \left(a_{1}+a_{2}+a_{3}+a_{4}\right)\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}}\right) \text { is }

Option: 1

4


Option: 2

16


Option: 3

3


Option: 4

None of these


Answers (1)

best_answer

\text { Apply } A M \geqslant G M \text { on } a_{1}, a_{2}, a_{3}, a_{4}

\begin{aligned} &\frac{a_{1}+a_{2}+a_{3}+a_{4}}{4} \geqslant\left(a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4}\right)^{1 / 4}\\ &\left.\Rightarrow a_{1}+a_{2}+a_{3}+a_{4} \geqslant 4(a_{1} \cdot a_{2} \cdot a_{3} \cdot a_{4}\right)^{1 / 4}---(i)\\\\\ &\text { Apply } A M \geqslant G M \text { on } \frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}}, \frac{1}{a_{4}}\\ &\frac{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}}}{4} \geqslant\left(\frac{1}{a_{1}} \cdot \frac{1}{a_{2}} \cdot \frac{1}{a_{3}} \cdot \frac{1}{a_{4}}\right)^{1 / 4} \end{aligned}

\\\Rightarrow \frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}} \geqslant 4\left(\frac{1}{a_{1}} \cdot \frac{1}{a_{2}} \cdot \frac{1}{a_{3}} \cdot \frac{1}{a_{4}}\right)^{1 / 4}-\text ----{ (ii) }\\\\\\(i)\times(i i)\\ \left(a_{1}+a_{2}+a_{3}+a_{4}\right) \left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}}\right) \geqslant 4.4 \cdot 1

So minimum value is 16

Posted by

Kuldeep Maurya

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