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In $\triangle ABC,a,b,c$ are the sides of triangle satisfying $a^4+b^4+c^4-4abc+1=0$


Find the value of $\frac{a^2+b^2+c^2}{S}$,where $S$ is area of the triangle $ABC$and find the value of $1+\frac{R}{r}$

where $R$ is the circumradius and $r$ is the inradius of the triangle $ABC$


My attempt:$a^4+b^4+c^4-4abc+1=0$

I expanded $(a+b+c)^4=a^4+4a^3b+4a^3c+6a^2b^2+6a^2c^2+12a^2bc+4ab^3+4ac^3+12ab^2c+12abc^2+4bc^3+4b^3c+6b^2c^2+b^4+c^4$

But this expression has got complicated and not seeming helpful and i do not know any other method to solve this question.

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HINT:

Using AM-GM inequality

$$\dfrac{a^4+b^4+c^4+1}4\ge\sqrt[4]{a^4b^4c^4\cdot1}$$

Can you take it from here?

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HINT: use that $$\cos(A)+\cos(B)+\cos(C)=1+\frac{r}{R}$$ holds

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