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Question Number 59627 by naka3546 last updated on 12/May/19

Answered by tanmay last updated on 12/May/19

$$\frac{3-1}{3}\left[(a+b+c)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})\right]$$$$=(\frac{b+c}{a^2}+\frac{c+a}{b^2}+\frac{a+b}{c^2}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-\frac{1}{3}(a+b+c)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})$$$$=\frac{b+c}{a^2}+\frac{c+a}{b^2}+\frac{a+b}{c^2}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{3}(a+b+c)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})$$$$letlefthandsideexpression=p$$$$righthandsideexoression=q$$$$wehavetoprovep-q=+ve$$$$otq-p=-ve$$$$now$$$$q=p+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{3}(a+b+c)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})$$$$q-p$$$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}⩾3{\left(\frac{1}{abc}\right)}^{\frac{1}{3}}$$$$\frac{a+b+c}{3}⩾{\left(abc\right)}^{\frac{1}{3}}$$$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}⩾3{\left(\frac{1}{a^2{b}^2{c}^2}\right)}^{\frac{1}{3}}$$$$(a+b+c)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})⩾3{\left(abc\right)}^{\frac{1}{3}}\times 3{\left(\frac{1}{a^2{b}^2{c}^2}\right)}^{\frac{1}{3}}$$$$so$$$$\frac{1}{3}(a+b+c)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})⩾3{\left(abc\right)}^{\frac{1}{3}}\times \frac{1}{{\left(abc\right)}^{\frac{2}{3}}}$$$$\frac{1}{3}(a+b+c)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})⩾\frac{3}{{\left(abc\right)}^{\frac{1}{3}}}$$$$now$$$$3{\left(\frac{1}{abc}\right)}^{\frac{1}{3}}-\frac{3}{{\left(abc\right)}^{\frac{1}{3}}}$$$$=0correctiondone$$$$sop=q$$$$$$$$$$$$$$$$...$$$$....$$$$$$$$$$

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