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MATHMAA

# 16th KVJ Solutions4

Here solution for 16th KVJ Solutions4 part a in detail . Please feel free to send your opinion

4Q) a) Let a, b,c be the length of the sides of a triangle and r be the in-radius. Show that:

r< $$\frac{a^{2}+b^{2}+c^{2}}{3(a+b+c)}$$.

Solution:

We know that for a $$\Delta ABC$$

r =$$\frac{\Delta}{S}$$, Where $$\Delta$$ is Area of triangle ABC and S is semi-perimeter.

2S = a + b + c  and $$\Delta = \frac{1}{2}bc sin(A)$$ =$$\frac{1}{2}ca sin(B)$$=$$\frac{1}{2}ab sin(C)$$ .

r=$$\frac{\Delta}{\frac{a+b+c}{2}}$$ then r(a+b+c)=$$2\Delta$$

$$2\Delta$$=bc sin(A)

=ca Sin(B)

=ab Sin(C)

$$6\Delta$$=bc Sin(A) + ca Sin(B) + ab Sin(C)

3r(a+b+c)=$$6\Delta$$

3r(a+b+c)=bc Sin(A) + ca Sin(B) + ab Sin(C) -------(1)

We know that maximum value of sin is 1

so we get

bc Sin(A) < bc

ca Sin(B) < ca

ab Sin(C) < ab

bc Sin(A) + ca Sin(B) + ab Sin(C) < bc + ca +ab

From (1) we get ,

3r(a+b+c) < bc + ca + ab ------------(2)

We know that G.M $$\leq$$ A.M

$$\sqrt{b^{2}c^{2}} \leq \frac{b^{2}+c^{2}}{2}$$

bc $$\leq \frac{b^{2}+c^{2}}{2}$$

Similarly we get

ca $$\leq \frac{c^{2}+a^{2}}{2}$$

ab $$\leq \frac{a^{2}+b^{2}}{2}$$

$$( bc + ca + ab) \leq \frac{2(a^{2}+b^{2}+c^{2})}{2}$$

bc + ca + ab $$\leq (a^{2}+b^{2}+c^{2})$$

From (2) we get,

3r(a+b+c) < bc + ca + ab $$\leq (a^{2}+b^{2}+c^{2})$$

3r(a+b+c) < $$a^{2}+b^{2}+c^{2}$$

r < $$\frac{a^{2}+b^{2}+c^{2}}{3(a+b+c)}$$

Hence it is proved.