# Difference between revisions of "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 $\frac{16}{5}$.

## 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 $\frac{16}{5}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 