Question

If $a,b,c,d$ be positive real numbers such that $a+b+c+d=\frac{4}{3}$, then the minimum value of $\sqrt{\frac{1}{b+c+d}}+\sqrt{\frac{1}{c+d+a}}+\sqrt{\frac{1}{d+a+b}}+\sqrt{\frac{1}{a+b+c}}$ is :

• A. $3$
• B. $6$
• C. $4$
• D. $8$

Answer 1

The correct option is C $4$

Case of Minimum arise when $a=b=c=d$

So, $a+b+c+d=4a=4b=4c=4d=\frac{4}{3}$

So, $a=b=c=d=\frac{1}{3}$

Thus,

$\sqrt{\frac{1}{a+b+c}}+\sqrt{\frac{1}{b+c+d}}+\sqrt{\frac{1}{c+d+a}}+\sqrt{\frac{1}{d+a+b}}=\frac{4}{\sqrt{3/3}}=\frac{4}{\sqrt{1}}=4$