InterviewSolution
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InterviewSolution
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If a,b,c and d are four positive real numbers such that abcd=1 , what is the minimum value of `(1+a)(1+b)(1+c)(1+d)`. |
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Answer» Here, we are given `abcd = 1`. Now, `(1+a)(1+b)(1+c)(1+d) = 1+a+b+c+d+(ab+bc+cd+ad+bd+ac)+(abc+bcd+cda+dab)+abcd` `=1+a+b+c+d+(ab+bc+ac+1/(ab)+1/(bc)+1/(cd))+(1/a+1/b+1/c+1/d)+1`...(As abcd = 1) `=1+1+(a+1/a)+(b+1/b)+(c+1/c)+(d+1/d)+(ab+1/(ab))+(bc+1/(bc))+(ac+1/(ac))` Now, we know minimum value of `x+1/x` is `2`. `:. (1+a)(1+b)(1+c)(1+d) ge 1+1+2+2+2+2+2+2+2` `=> (1+a)(1+b)(1+c)(1+d) ge 16` `:.` Minimum value of given expression is `16`. |
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