Maximize $\sum\limits_{\mathrm{cyc}} a$ for $a^2+b^2+c^2+d^2+a+3b+5c+7d=4$

Question

If $a,b,c,d$ are real numbers such that $a^2+b^2+c^2+d^2+a+3b+5c+7d=4$, find the maximal value of $a+b+c+d$.

Hi there, I found this problem in an old Serbian contest archive. It's dated to $2002$, but I can't find any solution online or in the archive (archive is, inconveniently, only problems). I'm not sure where to start here as they're not positive real numbers so I can't use some of the more popular inequalities and it's also really puzzling on how to find a connection between $a^2+b^2+c^2+d^2$ and $a+b+c+d$ as the numbers can be in between $-1,0,1$. I'd really appreciate some help here, thanks!

Answer 1

first make whole squares of all terms. Add $\frac{1}{4},\frac{9}{4},\frac{25}{4},\frac{49}{4}$

So you get $$ ( a^2+2(\frac{1}{2})a+(\frac{1}{2})^2 ) + ( b^2+2(\frac{3}{2})b+(\frac{3}{2})^2 )+ ( c^2+2(\frac{5}{2})c+(\frac{5}{2})^2 )+( d^2+2(\frac{7}{2})d+(\frac{7}{2})^2 ) $$$$= 4+\frac{1}{4}+\frac{9}{4}+\frac{25}{4}+\frac{49}{4}$$

this implies

$$(a+\frac{1}{2})^2+(b+\frac{3}{2})^2+(c+\frac{5}{2})^2+(d+\frac{7}{2})^2 = 25$$

Now apply Cauchy-Schwarz inequality,

So

$${(a+\frac{1}{2})+(b+\frac{3}{2})+(c+\frac{5}{2})+(d+\frac{7}{2})} \leq \sqrt{{(a+\frac{1}{2})^2+(b+\frac{3}{2})^2+(c+\frac{5}{2})^2+(d+\frac{7}{2})^2}}*\sqrt{4}$$

Solving gives $$a+b+c+d \leq 2$$

Answer 2

Remark: Here is a solution which gives not only the maximum but also the minimum. To find them, we simply complete the square of quadratic functions.

Answer: The maximum of $a + b + c + d$ is $2$ when $a = 2, b = 1, c = 0, d = -1$.

The minimum of $a + b + c + d$ is $-18$ when $a = -3, b = -4, c = -5, d = -6$.

Proof.

Let $x = a + b + c + d$.

Using $d = x - a - b - c$, we have \begin{align*} &a^2 + b^2 + c^2 + d^2 + a + 3b + 5c + 7d - 4\\ ={}& 2\,{a}^{2}+ \left( 2\,b+2\,c-2\,x-6 \right) a \\ &\qquad +2\,{b}^{2}+2\,bc-2\,xb+2 \,{c}^{2}-2\,xc+{x}^{2}-4\,b-2\,c+7\,x-4\\[6pt] ={}& 2\left(a - \frac{-b - c + x + 3}{2} \right)^2 + \frac32\,{b}^{2}+ \left( -1+c-x \right) b \\[6pt] &\qquad +\frac32\,{c}^{2}+4\,x+c+\frac12\,{x}^{2} -xc- \frac{17}{2}\\[6pt] ={}& 2\left(a - \frac{-b - c + x + 3}{2} \right)^2 + \frac32\left(b - \frac{1 + x - c}{3}\right)^2\\[6pt] &\qquad + \frac{4}{3}\,{c}^{2}+ \left( -\frac23\,x+\frac43 \right) c+\frac{11}{3}\,x-{\frac {26}{3}}+\frac13 \,{x}^{2}\\[6pt] ={}& 2\left(a - \frac{-b - c + x + 3}{2} \right)^2 + \frac32\left(b - \frac{1 + x - c}{3}\right)^2 + \frac43\left(c - \frac{x-2}{4}\right)^2\\[6pt] &\qquad + \frac{(x + 18)(x - 2)}{4}. \tag{1} \end{align*} (Note: We complete the square in three variables ($a$, $b$ and $c$), one at a time.)

From (1), we have $$\frac{(x + 18)(x - 2)}{4} \le 0$$ which results in $-18 \le x \le 2$.

1. $x = 2$

From (1), we have $$a - \frac{-b - c + x + 3}{2} = b - \frac{1 + x - c}{3} = c - \frac{x-2}{4} = 0$$ which results in $a = 2, b = 1, c = 0$.

Thus, the maximum of $a + b + c + d$ is $2$ when $a = 2, b = 1, c = 0, d = -1$.

2. $x = -18$

From (1), we have $$a - \frac{-b - c + x + 3}{2} = b - \frac{1 + x - c}{3} = c - \frac{x-2}{4} = 0$$ which results in $a = -3, b = -4, c = -5$.

Thus, the minimum of $a + b + c + d$ is $-18$ when $a = -3, b = -4, c = -5, d = -6$.

We are done.

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$\phantom{2}$

Discussion.

Here we introduce a trick which I have used to prove this.

Once we motivate our solutions by some way, we can give the following simpler solutions.

We have \begin{align*} &2 - (a + b + c + d) + \frac15(a^2 + b^2 + c^2 + d^2 +a + 3b + 5c + 7d - 4)\\ ={}& \frac15(a - 2)^2 + \frac15(b - 1)^2 + \frac15c^2 + \frac15(d + 1)^2 \\ \ge{}& 0 \end{align*} which gives $a + b + c + d \le 2$.

On the other hand, we have \begin{align*} &a + b + c + d + 18 + \frac15(a^2 + b^2 + c^2 + d^2 +a + 3b + 5c + 7d - 4)\\ ={}& \frac15(a + 3)^2 + \frac15(b + 4)^2 + \frac15(c + 5)^2 + \frac15(d + 6)^2 \\ \ge{}& 0 \end{align*} which gives $a + b + c + d \ge -18$.