Maximization question using inequalities (Cauchy-Schwarz, AM-GM)

Question

Consider this system of equations:

\begin{equation*} a^2 + b^2 + c^2 + d^2 = 14 \\ 3a + 2b + c + d = 14 \end{equation*}

I want to find the maximum value of $d$ given that $a, b, c, d \in \mathbb{R}$. Here is some work I've done so far:

• By using the AM-GM inequality on the first equation, we have that $\frac{a^2 + b^2 + c^2 + d^2}{4} \geq \sqrt[4]{a^2b^2c^2d^2}$, and so $\frac{14}{4} \geq \sqrt{abcd}$, and so $abcd \leq \frac{49}{4}$.
• By using the AM-GM inequality on the second equation, we have that $\frac{3a + 2b + c + d}{4} \geq \sqrt[4]{6abcd}$, and so $\frac{14}{4} \geq \sqrt[4]{6abcd}$, and so $6abcd \leq \frac{2401}{16}$, and so $abcd \leq \frac{2401}{96}$. This doesn't give us any new information from the first bullet point.
• Then, I tried using the Cauchy-Schwarz Inequality. If we let $\hat{x} = (a, b, c, d), \hat{y} = (3, 2, 1, 1)$, then we have that $\hat{x} \cdot \hat{y} = 3a + 2b + c + d = 14 \leq ||x||||y|| = \sqrt{a^2 + b^2 + c^2 + d^2}\sqrt{3^2 + 2^2 + 1^2 + 1^2} = \sqrt{15(a^2 + b^2 + c^2 + d^2)} = \sqrt{15(14)} = \sqrt{210}$.
  However, we already know this anyways, as we are given that $3a + 2b + c + d = 14$, and $14 \leq \sqrt{210}$ already.

Can someone provide hints/a solution to this problem?

Answer 1

We can focus on $a,b,c$ to derive an inequality about $d$. That is , $$ (14-d)^2=(3a+2b+c)^2\leq(3^2+2^2+1^2)(a^2+b^2+c^2)=14\cdot(14-d^2).$$ Solve this inequality we have $d\leq\frac{28}{15}$, and the maximum can be achieved when the Cauchy-Schwarz Inequality takes $"="$ and the system of equations for $a,b,c,d$ hold, where $(a,b,c,d)=(39/15,26/15,13/15,28/15)$.

Answer 2

Hint: Esp since you don't know when equality holds, your AM-GM approaches aren't likely to work.
IE It seems very unlikely that the maximum is achieved when $ a = b = c = d$ or $ 3a = 2b = c = d$.
What I might believe is that $ 3a = 2b = c $ or $ a = b = c$ or $\frac{a}{3} = \frac{b}{2} = \frac{c}{1}$, and then $d$ is left to vary to maximize the leftover.

Hint: Try to form an inequality involving $d$ only. Your third idea involving CS is a great start, how can we isolate out $d$ terms?

$(14 -d )^2 = (3a + 2b + c)^2 \leq (a^2 + b^2 + c^2)(3^2 + 2^2 + 1^2) = (14-d^2) \times 14$