Question

If \[a,b,c,d\] are four positive real numbers such that \[abcd=1\], then the minimum value of \[\left( 1+a \right)\left( 1+b \right)\left( 1+c \right)\left( 1+d \right)\] is equal to

Answer 1

Hint: Expand the brackets and rearrange the terms using the condition that \[abcd=1\]. Use the Arithmetic Mean-Geometric Mean Property (AM-GM Property) which states that for any two positive real numbers \[x\] and \[y\], we have the condition that \[x+y\ge 2\sqrt{xy}\], to find the minimum value of the given expression.

Complete step-by-step answer:
We have four positive real numbers \[a,b,c,d\] such that \[abcd=1\]. We have to find the minimum value of the expression \[\left( 1+a \right)\left( 1+b \right)\left( 1+c \right)\left( 1+d \right)\].

We will begin by expanding the given expressions by multiplying the terms.

Thus, we have \[\left( 1+a \right)\left( 1+b \right)\left( 1+c \right)\left( 1+d \right)=\left( 1+a+b+ab \right)\left( 1+c \right)\left( 1+d \right)\].

Further simplifying the expression, we have \[\left( 1+a \right)\left( 1+b \right)\left( 1+c \right)\left( 1+d \right)=\left( 1+a+b+ab+c+ac+bc+abc \right)\left( 1+d \right)\].


Thus, we have \[\left( 1+a \right)\left( 1+b \right)\left( 1+c \right)\left( 1+d \right)=1+a+b+ab+c+ac+bc+abc+d+ad+bd+abd+cd+acd+bcd+abcd\]

We can rearrange to write it as \[\left( 1+a \right)\left( 1+b \right)\left( 1+c \right)\left( 1+d \right)=1+a+b+c+d+ab+ac+ad+bc+bd+cd+abc+abd+bcd+acd+abcd..\left( 1 \right)\]

We know that \[abcd=1\].

Thus, we can rearrange the terms to get \[abc=\dfrac{1}{d},abd=\dfrac{1}{c},bcd=\dfrac{1}{a},acd=\dfrac{1}{b}.....\left( 2 \right)\].

We can also rearrange the terms of equation \[abcd=1\] to get \[ab=\dfrac{1}{cd},ac=\dfrac{1}{bd},ad=\dfrac{1}{bc}.....\left( 3 \right)\].

Substituting the value \[abcd=1\] and equation \[\left( 2 \right),\left( 3 \right)\] in equation \[\left( 1 \right)\], we have \[\left( 1+a \right)\left( 1+b \right)\left( 1+c \right)\left( 1+d \right)=1+1+\left( a+\dfrac{1}{a} \right)+\left( b+\dfrac{1}{b} \right)+\left( c+\dfrac{1}{c} \right)+\left( d+\dfrac{1}{d} \right)+\left( cd+\dfrac{1}{cd} \right)+\left( bd+\dfrac{1}{bd} \right)+\left( bc+\dfrac{1}{bc} \right).....\left( 4 \right)\]

We will now use Arithmetic Mean-Geometric Mean Property (AM-GM Property) which says that for any two positive real numbers \[x\] and \[y\], we have the condition that \[x+y\ge 2\sqrt{xy}\].

Substituting \[x=a,y=\dfrac{1}{a}\] in the above equation, we have \[a+\dfrac{1}{a}\ge 2\sqrt{a\left( \dfrac{1}{a} \right)}=2\sqrt{1}=2\Rightarrow a+\dfrac{1}{a}\ge 2\].

Thus, for any two positive real numbers, we have \[a+\dfrac{1}{a}\ge 2\].

Hence, we have \[a+\dfrac{1}{a}\ge 2,b+\dfrac{1}{b}\ge 2,c+\dfrac{1}{c}\ge 2,d+\dfrac{1}{d}\ge 2,cd+\dfrac{1}{cd}\ge 2,bd+\dfrac{1}{bd}\ge 2,bc+\dfrac{1}{bc}\ge 2.....\left( 5 \right)\].

We also know that if \[x\ge a\] and \[y\ge b\] then \[x+y\ge a+b\].

Using the inequality of equation \[\left( 5 \right)\] in equation \[\left( 4 \right)\], we have \[\left( 1+a \right)\left( 1+b \right)\left( 1+c \right)\left( 1+d \right)\ge 2+2+2+2+2+2+2+2=16\].

Thus, we have the inequality \[\left( 1+a \right)\left( 1+b \right)\left( 1+c \right)\left( 1+d \right)\ge 16\].

Hence, the minimum value of expression \[\left( 1+a \right)\left( 1+b \right)\left( 1+c \right)\left( 1+d \right)\] is \[16\] which holds when all the four terms are equal, i.e., \[a=b=c=d=1\].


Note: We can also solve this question by observing that the expression \[\left( 1+a \right)\left( 1+b \right)\left( 1+c \right)\left( 1+d \right)\] will attain minimum value only when \[a=b=c=d\] and thus, we need to find the roots of the equation \[{{x}^{4}}=1\]. The only positive real roots of the equation is \[x=1\] and thus, get the minimum value.