If $$a,b,c\in R^+$$ such that $$a^2+b^2+c^2 = 1$$, find the minimum value of
$$S = \dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}-\dfrac{2(a^2+b^2+c^2)}{abc} $$

Question

If $$a,b,c\in R^+$$ such that $$a^2+b^2+c^2 = 1$$, find the minimum value of 
$$S = \dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}-\dfrac{2(a^2+b^2+c^2)}{abc}        $$

Uniteasy Admin · Accepted Answer

when $$a=b=c$$, $$a^2=\dfrac{1}{3}$$ 
$$S = \dfrac{3}{a^2}-6 =\dfrac{3}{\dfrac{1}{3} }-6 = 3  $$
Guess: $$S\geq 3 $$
$$\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}-3-\dfrac{2(a^3+b^3+c^3)}{abc} $$
$$=\dfrac{a^2+b^2+c^2}{a^2}+\dfrac{a^2+b^2+c^2}{b^2}+\dfrac{a^2+b^2+c^2}{c^2}-3-2\Big( \dfrac{a^2}{bc}+\dfrac{b^2}{ac}+\dfrac{c^2}{ab} \Big)    $$
$$=\dfrac{b^2+c^2}{a^2}+\dfrac{a^2+c^2}{b^2}+\dfrac{a^2+b^2}{c^2}-2\Big( \dfrac{a^2}{bc}+\dfrac{b^2}{ac}+\dfrac{c^2}{ab} \Big)     $$
 $$=a^2\Big( \dfrac{1}{b^2}+\dfrac{1}{c^2}\Big) +b^2\Big( \dfrac{1}{a^2}+\dfrac{1}{c^2}  \Big) +c^2\Big( {\dfrac{1}{a^2}+\dfrac{1}{b^2}  } \Big)-2\Big( \dfrac{a^2}{bc}+\dfrac{b^2}{ac}+\dfrac{c^2}{ab} \Big)     $$ 
$$=a^2\Big( \dfrac{1}{b}-\dfrac{1}{c}  \Big) ^2 +b^2\Big( \dfrac{1}{a}-\dfrac{1}{c}  \Big) ^2+c^2\Big( \dfrac{1}{a}-\dfrac{1}{b}  \Big) ^2$$
$$\geq 0 $$
Therefore, the minimum of S is 3

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