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Question-59627

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Question Number 59627 by naka3546 last updated on 12/May/19
Answered by tanmay last updated on 12/May/19
((3−1)/3)[(a+b+c)((1/a^2 )+(1/b^2 )+(1/c^2 ))] =(((b+c)/a^2 )+((c+a)/b^2 )+((a+b)/c^2 )+(1/a)+(1/b)+(1/c))−(1/3)(a+b+c)((1/a^2 )+(1/b^2 )+(1/c^2 )) =((b+c)/a^2 )+((c+a)/b^2 )+((a+b)/c^2 )+(1/a)+(1/b)+(1/c)−(1/3)(a+b+c)((1/a^2 )+(1/b^2 )+(1/c^2 )) let left hand side expression=p right hand side exoression=q we have to prove p−q=+ve ot q−p=−ve now q=p+(1/a)+(1/b)+(1/c)−(1/3)(a+b+c)((1/a^2 )+(1/b^2 )+(1/c^2 )) q−p (1/a)+(1/b)+(1/c)≥3((1/(abc)))^(1/3) ((a+b+c)/3)≥(abc)^(1/3) (1/a^2 )+(1/b^2 )+(1/c^2 )≥3((1/(a^2 b^2 c^2 )))^(1/3) (a+b+c)((1/a^2 )+(1/b^2 )+(1/c^2 ))≥3(abc)^(1/3) ×3((1/(a^2 b^2 c^2 )))^(1/3) so (1/3)(a+b+c)((1/a^2 )+(1/b^2 )+(1/c^2 ))≥3(abc)^(1/3) ×(1/((abc)^(2/3) )) (1/3)(a+b+c)((1/a^2 )+(1/b^2 )+(1/c^2 ))≥(3/((abc)^(1/3) )) now 3((1/(abc)))^(1/3) −(3/((abc)^(1/3) )) =0 correction done so p=q ... ....
3−13[(a+b+c)(1a2+1b2+1c2)]=(b+ca2+c+ab2+a+bc2+1a+1b+1c)−13(a+b+c)(1a2+1b2+1c2)=b+ca2+c+ab2+a+bc2+1a+1b+1c−13(a+b+c)(1a2+1b2+1c2)letlefthandsideexpression=prighthandsideexoression=qwehavetoprovep−q=+veotq−p=−venowq=p+1a+1b+1c−13(a+b+c)(1a2+1b2+1c2)q−p1a+1b+1c⩾3(1abc)13a+b+c3⩾(abc)131a2+1b2+1c2⩾3(1a2b2c2)13(a+b+c)(1a2+1b2+1c2)⩾3(abc)13×3(1a2b2c2)13so13(a+b+c)(1a2+1b2+1c2)⩾3(abc)13×1(abc)2313(a+b+c)(1a2+1b2+1c2)⩾3(abc)13now3(1abc)13−3(abc)13=0correctiondonesop=q…….

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