Minimum $e$ where $a,b,c,d,e$ are reals such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$

Question

I have a question about this 1978 USAMO problem:

Given that $a,b,c,d,e$ are real numbers such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$, find the maximum value that $e$ can attain.

I had the following solution: Let $a+b+c+d=x$. Then $x+e=8\implies e=8-x$. Also, $\frac{a^2+1}{2}+\frac{b^2+1}{2}+\frac{c^2+1}{2}+\frac{d^2+1}{2}\geq (a+b+c+d)=x$ by AM-GM inequality. Hence, $a^2+b^2+c^2+d^2\geq 2x-4$.

Now we have $2x-4+e^2\leq a^2+b^2+c^2+d^2+e^2=16$. Substituting $x=8-e$, we get $e^2-2e-4\leq 0$. We can easily calculate that the lowest value that $e$ can attain is $1 -\sqrt{5}$.

However, the answer given on the internet is $\frac{16}{5}$. Where am I going wrong?

EDIT $1$ -- Is this a case of how the value $1-\sqrt{5}$ can never be attained by $e$, although the inequality is true?

EDIT $2$ -- It seems that we need to find the maximum. By my method, I’ve found the maximum to be $1+\sqrt{5}$. This is greater than $\frac{16}{5}$. Have I found a sharper inequality?

Answer 1

I don't understand, wouldn't the minimum value be $0$? if $$a=b=c=d=2$$ and $$e=0$$ both equations are satisfied. Clearly, zero would also be the smallest positive number.

Answer 2

By the RMS-AM inequality:

$$ \dfrac{a^2+b^2+c^2+d^2}{4} \ge \left(\dfrac{a+b+c+d}{4}\right)^2 = \dfrac{(8-e)^2}{16} \tag{1} $$

Therefore:

$$ e^2 = 16-(a^2+b^2+c^2+d^2) \le 16 - \dfrac{(8-e)^2}{4} \;\;\iff\;\; e(5e-16) \le 0 \tag{2} $$

Equality is in fact attained for $\,e = \dfrac{16}{5}\,$ and $\,a=b=c=d\,$.

Where am I going wrong?

Besides miscopying the problem as a "minimum" rather than "maximum", the equality case in your first step ("by AM-GM") is attained for $\,a=b=c=d=1\,$, but that's not necessarily where the global maximum for $\,e\,$ is attained.

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[ EDIT ]  Condition $\,(2) \iff e \in \left[0, \,16/5\right]\,$, which establishes both the minimum $\,e=0\,$ and maximum $\,e=16/5\,$, both attained when the other variables are equal $\,a=b=c=d\,$ so that the RMS-AM inequality $\,(1)\,$ becomes an equality. This should answer both the question referenced in the title (about the maximum $e$) as well as the posted variation (about the minimum $e$).

Answer 3

Given $a+b+c+d+e=8$ and $a+b+c+d=8-e$ $$(8-e)^2=(a+b+c+d)^2$$ Now expand $(a+b+c+d)^2$ $$=a^2+b^2+c^2+d^2+2(ab+ac+ad+bc+bd+cd)$$ $$\le4(a^2+b^2+c^2+d^2)$$

And Now, $a^2+b^2+c^2+d^2+e^2=16$

$a^2+b^2+c^2+d^2=16-e^2$ $$(16-e^2)=a^2+b^2+c^2+d^2$$$$4(a^2+b^2+c^2+d^2)\ge(a+b+c+d)^2$$$$4(16-e^2)\ge(8-e)^2$$$$64-4e^2\ge64-16e^2+e^2$$$$5e^2-16e\le0$$$$e(5e-16)\le0$$$$e\le\dfrac{16}{5}$$

Answer 4

Intuitive approach:

For a given $$(a+b+c+d=k)$$

We want

$$Min(a^2+b^2+c^2+d^2)$$

In order to maximize $e$

This is true only when $a=b=c=d$

From that we have

$$4x+e=8,\; 4x^2+e^2=16$$

Giving $x=\frac{6}{5}$ and $e=\frac{16}{5}$

Leading to the maximum value