1978 USAMO Problems/Problem 1

Problem

Given that $a,b,c,d,e$ are real numbers such that

$a+b+c+d+e=8$ ,

$a^2+b^2+c^2+d^2+e^2=16$ .

Determine the maximum value of $e$ .

Solution 1

By Cauchy Schwarz, we can see that $(1+1+1+1)(a^2+b^2+c^2+d^2)\geq (a+b+c+d)^2$ thus $4(16-e^2)\geq (8-e)^2$ Finally, $e(5e-16) \geq 0$ which means $\frac{16}{5} \geq e \geq 0$ so the maximum value of $e$ is $\frac{16}{5}$ .

from: Image from Gon Mathcenter.net

Solution 2

Seeing as we have an inequality with constraints, we can use Lagrange multipliers to solve this problem. We get the following equations:

$(1)\hspace*{0.5cm} a+b+c+d+e=8\\ (2)\hspace*{0.5cm} a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16\\ (3)\hspace*{0.5cm} 0=\lambda+2a\mu\\ (4)\hspace*{0.5cm} 0=\lambda+2b\mu\\ (5)\hspace*{0.5cm} 0=\lambda+2c\mu\\ (6)\hspace*{0.5cm} 0=\lambda+2d\mu\\ (7)\hspace*{0.5cm} 1=\lambda+2e\mu$

If $\mu=0$ , then $\lambda=0$ according to $(6)$ and $\lambda=1$ according to $(7)$ , so $\mu \neq 0$ . Setting the right sides of $(3)$ and $(4)$ equal yields $\lambda+2a \mu= \lambda+2b \mu \implies 2a\mu=2b \mu \implies a=b$ . Similar steps yield that $a=b=c=d$ . Thus, $(1)$ becomes $4d+e=8$ and $(2)$ becomes $4d^{2}+e^{2}=16$ . Solving the system yields $e=0,\frac{16}{5}$ , so the maximum possible value of $e$ is $\frac{16}{5}$ .

Solution 3

A re-writing of Solution 1 to avoid the use of Cauchy Schwarz. We have $(a+b+c+d)^2=(8-e)^2,$ and $a^2+b^2+c^2+d^2=16-e^2.$ The second equation times 4, then minus the first equation, $(a-b)^2+(a-c)^2+(a-d)^2+(b-c)^2+(b-d)^2+(c-d)^2=4(16-e^2)-(8-e)^2.$ The rest follows.

J.Z.

Solution 4

By the Principle of Insufficient Reasons, since $a,b,c,d$ are indistinguishable variables, the maximum of $e$ is acheived when $a=b=c=d$ , so we have $4a+\max e=8$ $4a^2+(max e)^2=16$ $\implies e=\boxed{\frac{16}{5}}$ . $\square$ ~Ddk001

1978 USAMO (Problems • Resources)
Preceded by First Question	Followed by Problem 2
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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