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Let `agt0, bgt0, cgt0` and `a+b+c=6` then `((ab+1)^(2))/(b^(2))+((bc+1)^(2))/(c^(2))+((ca+1)^(2))/(a^(2))` may beA. `75/4`B. `35`C. `15`D. `10` |
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Answer» Correct Answer - A::B A.M. `ge` H.M for `1/a, 1/b, 1/c` we get `(1/a+1/b+1/c)/3 ge 3/6` `implies 1/a+1/b+1/c ge 3/2` Now, `((a+1/b)^(2)+(b+1/c)^(2)+(c+1/a)^(2))/3 ge ((a+1/b+b+1/c+c+1/a)/3)^(2) ge ((6+3/2)/3)^(2)` `(a+1/b)^(2)+(b+1/c)^(2)+(c+1/a)^(2) ge 75/4` |
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