if a,b,c,d are four positive real no.s such that abcd=1, what is the minimum value of (1+a)(1+b)(1+c)(1+d)
chaipanch98:
from which chapter. cause there are various methods.
real no.s class 10
do u know abt AM and GM inequality
am is always is always greater than or equal to gm??
yeah that one
yes
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Call X=(1+a)(1+b)(1+c)(1+d)
X=1+a+b+c+d+ab+ac+ad+bc+bd+cd+abc+bcd+acd+abd+abcdCollecting terms:=1+(a+b+c+d)+(ab+ac+ad+bc+bd+cd)+(abc+bcd+acd+abd)+abcdUsing abcd=1:X=1+(a+b+c+d)+(ab+ac+ad+1/ad+1/ac+1/ab)+(1/d+1/a+1/b+1/c)+1Rearranging:X=2+(a+1/a)+(b+1/b)+(c+1/c)+(d+1/d)+(ab+1/ab+ac+1/ac+ad+1/ad)Now using the identity [x+1/x≥2] for all x>0, and assuming a,b,c,d>0:
X=2+(a+1/a)+(b+1/b)+(c+1/c)+(d+1/d)+(ab+1/ab+ac+1/ac+ad+1/ad)=2+(>=2)+(>=2)+(>=2)+(>=2)+(>=2)+(>=2)+(>=2)≥16Hence, the MINIMUM value has to be 16, but on the max side, it can go to infinite.Call X=(1+a)(1+b)(1+c)(1+d)
X=1+a+b+c+d+ab+ac+ad+bc+bd+cd+abc+bcd+acd+abd+abcdCollecting terms:=1+(a+b+c+d)+(ab+ac+ad+bc+bd+cd)+(abc+bcd+acd+abd)+abcdUsing abcd=1:X=1+(a+b+c+d)+(ab+ac+ad+1/ad+1/ac+1/ab)+(1/d+1/a+1/b+1/c)+1Rearranging:X=2+(a+1/a)+(b+1/b)+(c+1/c)+(d+1/d)+(ab+1/ab+ac+1/ac+ad+1/ad)Now using the identity [x+1/x≥2] for all x>0, and assuming a,b,c,d>0:
X=2+(a+1/a)+(b+1/b)+(c+1/c)+(d+1/d)+(ab+1/ab+ac+1/ac+ad+1/ad)=2+(>=2)+(>=2)+(>=2)+(>=2)+(>=2)+(>=2)+(>=2)≥16Hence, the MINIMUM value has to be 16, but on the max side, it can go to infinite.
X=1+a+b+c+d+ab+ac+ad+bc+bd+cd+abc+bcd+acd+abd+abcdCollecting terms:=1+(a+b+c+d)+(ab+ac+ad+bc+bd+cd)+(abc+bcd+acd+abd)+abcdUsing abcd=1:X=1+(a+b+c+d)+(ab+ac+ad+1/ad+1/ac+1/ab)+(1/d+1/a+1/b+1/c)+1Rearranging:X=2+(a+1/a)+(b+1/b)+(c+1/c)+(d+1/d)+(ab+1/ab+ac+1/ac+ad+1/ad)Now using the identity [x+1/x≥2] for all x>0, and assuming a,b,c,d>0:
X=2+(a+1/a)+(b+1/b)+(c+1/c)+(d+1/d)+(ab+1/ab+ac+1/ac+ad+1/ad)=2+(>=2)+(>=2)+(>=2)+(>=2)+(>=2)+(>=2)+(>=2)≥16Hence, the MINIMUM value has to be 16, but on the max side, it can go to infinite.Call X=(1+a)(1+b)(1+c)(1+d)
X=1+a+b+c+d+ab+ac+ad+bc+bd+cd+abc+bcd+acd+abd+abcdCollecting terms:=1+(a+b+c+d)+(ab+ac+ad+bc+bd+cd)+(abc+bcd+acd+abd)+abcdUsing abcd=1:X=1+(a+b+c+d)+(ab+ac+ad+1/ad+1/ac+1/ab)+(1/d+1/a+1/b+1/c)+1Rearranging:X=2+(a+1/a)+(b+1/b)+(c+1/c)+(d+1/d)+(ab+1/ab+ac+1/ac+ad+1/ad)Now using the identity [x+1/x≥2] for all x>0, and assuming a,b,c,d>0:
X=2+(a+1/a)+(b+1/b)+(c+1/c)+(d+1/d)+(ab+1/ab+ac+1/ac+ad+1/ad)=2+(>=2)+(>=2)+(>=2)+(>=2)+(>=2)+(>=2)+(>=2)≥16Hence, the MINIMUM value has to be 16, but on the max side, it can go to infinite.
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