Statement I In any ∆ ABC, the minimum value r1+r2+r3r is 9.

We know that 1r1+1r2+1r3=1r and AM≥HMr1+r2+r33≥31r1+1r2+1r3r1+r2+r33≥31r r1+r2+r3r≥9·: Minimum value of r1+r2+r3r is 9, Statement I is true. Statement II is also true, but not the correct explanation of Statement I.

Statement I In any ∆ ABC, the minimum value r1+r2+r3r is 9.

We know that 1r1+1r2+1r3=1r and AM≥HMr1+r2+r33≥31r1+1r2+1r3r1+r2+r33≥31r r1+r2+r3r≥9·: Minimum value of r1+r2+r3r is 9, Statement I is true. Statement II is also true, but not the correct explanation of Statement I.

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Statement I In any $∆$ ABC, the minimum value $\frac{r_{1} + r_{2} + r_{3}}{r}$ is 9.

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Both Statement I and Statement I l are correct and Statement 11 is the correct expl anation of Statement I

Both Statement I and Statement II are correct but Statement It is not the correct explanation of Statement I

Statement I is correct but Statement 11 is incorrect

. Statement IJ is correct but Statement I is incorrect

answer is B.

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Detailed Solution

We know that

$\begin{aligned} \frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}} = \frac{1}{r} \\ and A M \geq H M \\ \frac{r_{1} + r_{2} + r_{3}}{3} \geq \frac{3}{\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}}} \\ \frac{r_{1} + r_{2} + r_{3}}{3} \geq \frac{3}{\frac{1}{r}} \end{aligned}$