Courses
Courses for Kids
Free study material
Free LIVE classes
More

# 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

Last updated date: 21st Mar 2023
Total views: 305.1k
Views today: 2.83k
Verified
305.1k+ views
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.

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.