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

Accordting to Cauchy-Schwarz Inequalities, we can see $(1+1+1+1)(a^2+b^2+c^2+d^2)\geqslant (a+b+c+d)^2$ thus, $4(16-e^2)\geqslant (8-e)^2$ Finally, $e(5e-16) \geqslant 0$ that mean, $\frac{16}{5} \geqslant e \geqslant 0$ 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: $\newline(1) a+b+c+d+e=8\newline (2) a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16\newline (3) 0=\lambda+2a\mu\newline (4) 0=\lambda+2b\mu\newline (5) 0=\lambda+2c\mu\newline (6) 0=\lambda+2d\mu\newline (7) 1=\lambda+2e\mu\newline$ 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 that maximum possible value of $e$ is $\frac{16}{5}$

1978 USAMO (Problems • Resources)
Preceded by First Question	Followed by Problem 2
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1978 USAMO Problems/Problem 1

Contents

Problem

Solution 1

Solution 2

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