No, the answer is not 2.500

Algebra Level 5

$\begin{aligned} &&\sqrt{ \dfrac a{b+c+d+e}} + \sqrt{ \dfrac b{a+c+d+e}} + \sqrt{ \dfrac c{a+b+d+e}} \\ && + \sqrt{ \dfrac d{a+b+c+e}} + \sqrt{ \dfrac e{a+b+c+d}} \end{aligned}$

Over all non-negative real numbers $a,b,c,d$ and $e$ , find the infimium value of the expression above.

Give your answer to 3 decimal places.

The answer is 2.000.

1 solution

Ariel Gershon
Mar 18, 2016

Let $n=a+b+c+d+e$ . If we call the expression above $P$ , then $P = \sum_{k=a,b,c,d,e}\sqrt{\frac{k}{n-k}}$ Using a little algebraic manipulation we get the following: $0 \ge -k(2k-n)^2 = 4k^2(n-k)-kn^2$ $kn^2 \ge 4k^2(n-k)$ $\frac{k}{n-k} \ge \frac{4k^2}{n^2}$ Note that in order for $P$ to be defined, $n-k \neq 0$ . Since each $k$ is nonnegative, it follows that both $n-k$ and $n$ are strictly positive, hence the above maneuver is justified. Now taking the square root of both sides gives: $\sqrt{\frac{k}{n-k}} \ge \frac{2k}{n}$

Therefore, $P \ge \sum_{k} \frac{2k}{n} = \frac{2n}{n} = 2$

Since the problem says $a,b,c,d,e$ are nonnegative real numbers, we can select $a=b=1, c=d=e=0$ which gives the minimum value $\boxed{2}$ .

Sorry, I have no idea how to use algebraic manipulation get the equation?

Kris Chen - 5 years, 2 months ago

In the algebraic manipulation part, I think it's easier for the others to follow the solution if you write like this: $\sqrt{\frac{k}{n-k}} \geq \frac{2k}{n} \Leftrightarrow kn^2 \geq 4k^2(n-k) \Leftrightarrow 0 \geq 4k^2(n-k)-kn^2 \Leftrightarrow 0 \geq -k(2k-n)^2$ .

This is true for all positive $k$ .

Tran Quoc Dat - 5 years, 2 months ago

No, the answer is not 2.500

The answer is 2.000.

1 solution

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