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}

Collected in the board: Inequality

Steven Zheng posted 1 month ago


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


Steven Zheng posted 1 month ago

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