An algebra problem by Anish Roy

Algebra Level 4

If $a$ , $b$ , $c$ , and $d$ are positive real numbers such that $abcd = 16$ . Then find the minimum value of $(1+a)(1+b)(1+c)(1+d)$

The answer is 81.

2 solutions

Marco Brezzi
Aug 5, 2017

$\textbf{Claim:}$

The minimum is $81$ and occurs when $a=b=c=d=2$

$\textbf{Proof:}$

Suppose the claim wasn't true and one of the variables, wlog $a$ , was $a=2+k$ where $k$ is a non-zero amount and $k>-2$ so that $a$ is still positive. To balance this, we will set $b=\dfrac{4}{2+k}$ and fix the other two variables to $2$

If we now calculate $P=(a+1)(b+1)(c+1)(d+1)$ we get

$P=(k+3)\left(\dfrac{k+6}{k+2}\right)\cdot 3 \cdot 3$

We will now just need to prove that the part depending on $k$ is $>9$

$(k+3)\left(\dfrac{k+6}{k+2}\right)>9$

$(k+3)(k+6)>9(k+2)$

$k^2+9k+18>9k+18 \iff k^2>0$

Since $k\neq 0$ , this is always true. Follows $P>81$

$QED$

I don't think it is a proper solution to the problem. After coming up with such an idea, we should first verify that the minimum can be achieved. In this case I don't think it can be, considering the fact that in the 4 inequalities above equality only holds when all of the variables are 1, which would mean that their product is 1 instead of 16. This is clearly a contradiction.

Sándor Daróczi - 3 years, 10 months ago

Thanks for pointing out my error, I edited my solution. I am now going to also post it as a report since $64$ is not the right answer

Marco Brezzi - 3 years, 10 months ago

U did the same, but the proof can be done with AM-GM more easily

Md Zuhair - 3 years, 10 months ago

My first step is just the basic idea to then construct $AM-GM$ with induction. Quoting that inequality I would have probably just saved some lines of proof, making it slightly quicker, but not easier, the concept is the same

Marco Brezzi - 3 years, 10 months ago

I still have got concerns about the validity of your solution. What happens when the other 2 variables do not have the fix value of 2? You can not just intuitively assume such claims. With this solution you have just shown that if $ab=4$ we have $(a+1)(b+1) \geq 9$ with the equality case $a=2$ and $b=2$ .

Sándor Daróczi - 3 years, 10 months ago

Since I am not allowed to write a solution, I'll leave here a sketch about how I tackled this problem, just for indication. My solution might not be among the most beautiful and straightforward approaches, but I think it is worth considering.

If we expand the product $(1+a)(1+b)(1+c)(1+d)$ and employ AM-GM separately to the terms of the same degree we obtain

$(1+a)(1+b)(1+c)(1+d) = 1 + a+b+c+d+ab+ac+ad+bc+bd+cd+abc+abd+acd+bcd+abcd$

1.: $a+b+c+d \geq 4 \sqrt[4]{abcd} = 8$

2.: $ab+ac+ad+bc+bd+cd \geq 6 \sqrt[6]{(abcd)^3} = 24$

3.: $abc+abd+acd+bcd \geq 4 \sqrt[4]{(abcd)^3} = 32$

4.: $abcd = 16$

Finally:

$(1+a)(1+b)(1+c)(1+d) = 1 + a+b+c+d+ab+ac+ad+bc+bd+cd+abc+abd+acd+bcd+abcd \geq 1+8+24+32+16 = 81$ .

Equality holds if $a=b=c=d=2$ .

Sándor Daróczi - 3 years, 10 months ago

Nice solution. It may not seem a very elegant approach but I think it's one of the best ones, specially to generalize for a case with $a_1\cdot a_2 \cdots a_n = k>0$

Marco Brezzi - 3 years, 10 months ago

sorry guys you are right on your solution. I should not have written positive integers, i have made necessary corrections and please have a look at my solution.

Anish Roy - 3 years, 10 months ago

Chew-Seong Cheong
Aug 12, 2017

By Hölder's inequality , we have:

$\begin{aligned} (1+a)^\frac 14 (1+b)^\frac 14 (1+c)^\frac 14 (1+d)^\frac 14 & \ge 1^\frac 14 + (abcd)^\frac 14 = 1 + 2 = 3 \\ \implies (1+a) (1+b) (1+c) (1+d) & \ge 3^4 = \boxed{81} \end{aligned}$ .

An algebra problem by Anish Roy

The answer is 81.

2 solutions

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